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11
Kinematics of Particles
The motion of the paraglider can be described in terms of its
position, velocity, and acceleration. When landing, the pilot of the
paraglider needs to consider the wind velocity and the relative
motion of the glider with respect to the wind. The study of
motion is known as kinematics and is the subject of this chapter.
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Kinematics of Particles
Objectives
Introduction
11.1
RECTILINEAR MOTION OF
PARTICLES
11.1A Position, Velocity, and
Acceleration
11.1B Determining the Motion of a
Particle
11.2 SPECIAL CASES AND
RELATIVE MOTION
11.2A Uniform Rectilinear Motion
11.2B Uniformly Accelerated
Rectilinear Motion
11.2C Motion of Several Particles
*11.3 GRAPHICAL SOLUTIONS
11.4 CURVILINEAR MOTION OF
PARTICLES
11.4A Position, Velocity, and
Acceleration Vectors
11.4B Derivatives of Vector Functions
11.4C Rectangular Components of
Velocity and Acceleration
11.4D Motion Relative to a Frame in
Translation
11.5
NON-RECTANGULAR
COMPONENTS
11.5A Tangential and Normal
Components
11.5B Radial and Transverse
Components
• Describe the basic kinematic relationships between
position, velocity, acceleration, and time.
• Solve problems using these basic kinematic
relationships and calculus or graphical methods.
• Define position, velocity, and acceleration in terms of
Cartesian, tangential and normal, and radial and
transverse coordinates.
• Analyze the relative motion of multiple particles by
using a translating coordinate system.
• Determine the motion of a particle that depends on
the motion of another particle.
• Determine which coordinate system is most appropriate for solving a curvilinear kinematics problem.
• Calculate the position, velocity, and acceleration of a
particle undergoing curvilinear motion using Cartesian,
tangential and normal, and radial and transverse
coordinates.
Introduction
Chapters 1 to 10 were devoted to statics, i.e., to the analysis of bodies at
rest. We now begin the study of dynamics, which is the part of mechanics
that deals with the analysis of bodies in motion.
Although the study of statics goes back to the time of the Greek
philosophers, the first significant contribution to dynamics was made by
Galileo (1564–1642). Galileo’s experiments on uniformly accelerated bodies led Newton (1642–1727) to formulate his fundamental laws of motion.
Dynamics includes two broad areas of study:
1. Kinematics, which is the study of the geometry of motion. The principles
of kinematics relate the displacement, velocity, acceleration, and time
of a body’s motion, without reference to the cause of the motion.
2. Kinetics, which is the study of the relation between the forces acting
on a body, the mass of the body, and the motion of the body. We use
kinetics to predict the motion caused by given forces or to determine
the forces required to produce a given motion.
Chapters 11 through 14 describe the dynamics of particles; in
Chap. 11, we consider the kinematics of particles. The use of the word
particles does not mean that our study is restricted to small objects; rather,
it indicates that in these first chapters we study the motion of bodies—
possibly as large as cars, rockets, or airplanes—without regard to their size
or shape. By saying that we analyze the bodies as particles, we mean that
we consider only their motion as an entire unit; we neglect any rotation
about their own centers of mass. In some cases, however, such a rotation is
not negligible, and we cannot treat the bodies as particles. Such motions are
analyzed in later chapters dealing with the dynamics of rigid bodies.
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11.1
Rectilinear Motion of Particles
617
In the first part of Chap. 11, we describe the rectilinear motion of
a particle; that is, we determine the position, velocity, and acceleration of
a particle at every instant as it moves along a straight line. We first use
general methods of analysis to study the motion of a particle; we then
consider two important particular cases, namely, the uniform motion and
the uniformly accelerated motion of a particle (Sec. 11.2). We then discuss
the simultaneous motion of several particles and introduce the concept
of the relative motion of one particle with respect to another. The first part
of this chapter concludes with a study of graphical methods of analysis
and their application to the solution of problems involving the rectilinear
motion of particles.
In the second part of this chapter, we analyze the motion of a particle as it moves along a curved path. We define the position, velocity, and
acceleration of a particle as vector quantities and introduce the derivative
of a vector function to add to our mathematical tools. We consider applications in which we define the motion of a particle by the rectangular components of its velocity and acceleration; at this point, we analyze the
motion of a projectile (Sec. 11.4C). Then we examine the motion of a
particle relative to a reference frame in translation. Finally, we analyze
the curvilinear motion of a particle in terms of components other than
rectangular. In Sec. 11.5, we introduce the tangential and normal components of an object’s velocity and acceleration and then examine the radial
and transverse components.
11.1
RECTILINEAR MOTION
OF PARTICLES
A particle moving along a straight line is said to be in rectilinear motion.
The only variables we need to describe this motion are the time, t, and
the distance along the line, x, as a function of time. With these variables,
we can define the particle’s position, velocity, and acceleration, which
completely describe the particle’s motion. When we study the motion of
a particle moving in a plane (two dimensions) or in space (three dimensions),
we will use a more general position vector rather than simply the distance
along a line.
11.1A
Position, Velocity, and
Acceleration
At any given instant t, a particle in rectilinear motion occupies some
position on the straight line. To define the particle’s position P, we choose
a fixed origin O on the straight line and a positive direction along the line.
We measure the distance x from O to P and record it with a plus or minus
sign, according to whether we reach P from O by moving along the line
in the positive or negative direction. The distance x, with the appropriate
sign, completely defines the position of the particle; it is called the position
coordinate of the particle. For example, the position coordinate
corresponding to P in Fig. 11.1a is x 5 15 m; the coordinate corresponding
to P9 in Fig. 11.1b is x9 5 22 m.
bee87342_ch11_615-717.indd 617
O
P
x
x
(a)
P'
1m
O
x
x'
(b)
1m
Fig. 11.1 Position is measured from a fixed
origin. (a) A positive position coordinate;
(b) a negative position coordinate.
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Kinematics of Particles
P
x
∆x
P'
O
(t) (t + ∆t)
x
Fig. 11.2 A small displacement Dx from
time t to time t 1 Dt.
When we know the position coordinate x of a particle for every value
of time t, we say that the motion of the particle is known. We can provide
a “timetable” of the motion in the form of an equation in x and t, such as
x 5 6t2 2 t3, or in the form of a graph of x versus t, as shown in Fig. 11.6.
The units most often used to measure the position coordinate x are the
meter (m) in the SI system of units† and the foot (ft) in the U.S. customary
system of units. Time t is usually measured in seconds (s).
Now consider the position P occupied by the particle at time t and
the corresponding coordinate x (Fig. 11.2). Consider also the position P9
occupied by the particle at a later time t 1 Dt. We can obtain the position
coordinate of P9 by adding the small displacement Dx to the coordinate x
of P. This displacement is positive or negative according to whether P9 is
to the right or to the left of P. We define the average velocity of the
particle over the time interval Dt as the quotient of the displacement Dx
and the time interval Dt as
Average velocity 5
Photo 11.1
The motion of this solar car can
be described by its position, velocity, and
acceleration.
Dx
Dt
If we use SI units, Dx is expressed in meters and Dt in seconds; the
average velocity is then expressed in meters per second (m/s). If we use
U.S. customary units, Dx is expressed in feet and Dt in seconds; the
average velocity is then expressed in feet per second (ft/s).
We can determine the instantaneous velocity v of a particle at the
instant t by allowing the time interval Dt to become infinitesimally small. Thus,
Instantaneous velocity 5 v 5 lim
Dt y0
Dx
Dt
The instantaneous velocity is also expressed in m/s or ft/s. Observing that
the limit of the quotient is equal, by definition, to the derivative of x with
respect to t, we have
Velocity of a particle
along a line
P
v5
v>0
x
(a)
v<0
P
x
(b)
Fig. 11.3
In rectilinear motion, velocity can
be only (a) positive or (b) negative along the
line.
d
dx
dt
We represent the velocity v by an algebraic number that can be positive or
negative.‡ A positive value of v indicates that x increases, i.e., that the particle
moves in the positive direction (Fig. 11.3a). A negative value of v indicates
that x decreases, i.e., that the particle moves in the negative direction
(Fig. 11.3b). The magnitude of v is known as the speed of the particle.
Consider the velocity v of the particle at time t and also its velocity
v 1 Dv at a later time t 1 Dt (Fig. 11.4). We define the average acceleration
of the particle over the time interval Dt as the quotient of Dv and Dt as
Average acceleration 5
P
(t)
v
P'
(t + ∆t)
Fig. 11.4
v + ∆v
Dv
Dt
†
x
A change in velocity from v to
v 1 Dv corresponding to a change in time
from t to t 1 Dt.
bee87342_ch11_615-717.indd 618
(11.1)
See Sec. 1.3.
As you will see in Sec. 11.4A, velocity is actually a vector quantity. However, since we are
considering here the rectilinear motion of a particle where the velocity has a known and fixed
direction, we need only specify its sense and magnitude. We can do this conveniently by using
a scalar quantity with a plus or minus sign. This is also true of the acceleration of a particle
in rectilinear motion.
‡
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11.1
Rectilinear Motion of Particles
619
If we use SI units, Dv is expressed in m/s and Dt in seconds; the average
acceleration is then expressed in m/s2. If we use U.S. customary units, Dv
is expressed in ft/s and Dt in seconds; the average acceleration is then
expressed in ft/s2.
We obtain the instantaneous acceleration a of the particle at the
instant t by again allowing the time interval Dt to approach zero. Thus,
Instantaneous acceleration 5 a 5 lim
Dt y0
Dv
Dt
The instantaneous acceleration is also expressed in m/s2 or ft/s2. The limit
of the quotient, which is by definition the derivative of v with respect to t,
measures the rate of change of the velocity. We have
Acceleration of a
particle along a line
a5
d
dv
dt
(11.2)
d 2x
dt 2
(11.3)
or substituting for v from Eq. (11.1),
a5
We represent the acceleration a by an algebraic number that can be positive or negative (see the footnote on the preceding page). A positive value
of a indicates that the velocity (i.e., the algebraic number v) increases.
This may mean that the particle is moving faster in the positive direction
(Fig. 11.5a) or that it is moving more slowly in the negative direction
(Fig. 11.5b); in both cases, Dv is positive. A negative value of a indicates
that the velocity decreases; either the particle is moving more slowly in
the positive direction (Fig. 11.5c), or it is moving faster in the negative
direction (Fig. 11.5d).
Sometimes we use the term deceleration to refer to a when the speed
of the particle (i.e., the magnitude of v) decreases; the particle is then moving
more slowly. For example, the particle of Fig. 11.5 is decelerating in parts
b and c; it is truly accelerating (i.e., moving faster) in parts a and d.
v'
v'
v
P'
P
v
P'
x
P
x
a>0
a>0
(b)
(a)
v
P
v'
v'
P'
v
P'
P
x
x
a<0
a<0
(c)
(d)
Fig. 11.5
Velocity and acceleration can be in the same or different directions.
(a, d) When a and v are in the same direction, the particle speeds up;
(b, c) when a and v are in opposite directions, the particle slows down.
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Kinematics of Particles
We can obtain another expression for the acceleration by eliminating
the differential dt in Eqs. (11.1) and (11.2). Solving Eq. (11.1) for dt, we
have dt 5 dx/v; substituting into Eq. (11.2) gives us
a5v
d
dv
d
dx
Concept Application 11.1
x (m)
32
Consider a particle moving in a straight line, and assume that its position
is defined by
24
x 5 6t2 2 t3
16
where t is in seconds and x in meters. We can obtain the velocity v at any
time t by differentiating x with respect to t as
8
0
2
4
v5
6
t (s)
v (m/s)
12
0
4
6
2
t (s)
–24
–36
a (m/s2)
2
4
6
0
–12
–24
Fig. 11.6 Graphs of position,
velocity, and acceleration as
functions of time for Concept
Application 11.1.
bee87342_ch11_615-717.indd 620
dx
5 12t 2 3t 2
dt
We can obtain the acceleration a by differentiating again with respect to t.
Hence,
a5
–12
12
(11.4)
t (s)
dv
5 12 2 6t
dt
In Fig. 11.6, we have plotted the position coordinate, the velocity, and the
acceleration. These curves are known as motion curves. Keep in mind,
however, that the particle does not move along any of these curves; the
particle moves in a straight line.
Since the derivative of a function measures the slope of the corresponding curve, the slope of the x–t curve at any given time is equal to
the value of v at that time. Similarly, the slope of the v–t curve is equal
to the value of a. Since a 5 0 at t 5 2 s, the slope of the v–t curve must
be zero at t 5 2 s; the velocity reaches a maximum at this instant. Also,
since v 5 0 at t 5 0 and at t 5 4 s, the tangent to the x–t curve must be
horizontal for both of these values of t.
A study of the three motion curves of Fig. 11.6 shows that the motion
of the particle from t 5 0 to t 5 ∞ can be divided into four phases:
1. The particle starts from the origin, x 5 0, with no velocity but with
a positive acceleration. Under this acceleration, the particle gains a
positive velocity and moves in the positive direction. From t 5 0 to
t 5 2 s, x, v, and a are all positive.
2. At t 5 2 s, the acceleration is zero; the velocity has reached its
maximum value. From t 5 2 s to t 5 4 s, v is positive, but a is
negative. The particle still moves in the positive direction but more
slowly; the particle is decelerating.
3. At t 5 4 s, the velocity is zero; the position coordinate x has reached
its maximum value (32 m). From then on, both v and a are negative;
the particle is accelerating and moves in the negative direction with
increasing speed.
4. At t 5 6 s, the particle passes through the origin; its coordinate x is
then zero, while the total distance traveled since the beginning of the
motion is 64 m (i.e., twice its maximum value). For values of t larger
than 6 s, x, v, and a are all negative. The particle keeps moving in
the negative direction—away from O—faster and faster. 12/10/15 11:36 AM
11.1
11.1B
Rectilinear Motion of Particles
621
Determining the Motion of a
Particle
We have just seen that the motion of a particle is said to be known if we
know its position for every value of the time t. In practice, however, a
motion is seldom defined by a relation between x and t. More often, the
conditions of the motion are specified by the type of acceleration that the
particle possesses. For example, a freely falling body has a constant
acceleration that is directed downward and equal to 9.81 m/s2 or 32.2 ft/
s2, a mass attached to a stretched spring has an acceleration proportional
to the instantaneous elongation of the spring measured from its equilibrium
position, etc. In general, we can express the acceleration of the particle
as a function of one or more of the variables x, v, and t. Thus, in order
to determine the position coordinate x in terms of t, we need to perform
two successive integrations.
Let us consider three common classes of motion.
1. a 5 f(t). The Acceleration Is a Given Function of t. Solving Eq. (11.2)
for dv and substituting f(t) for a, we have
dv 5 a dt
dv 5 f(t) dt
Integrating both sides of the equation, we obtain
e dv 5 e f(t) dt
This equation defines v in terms of t. Note, however, that an arbitrary
constant is introduced after the integration is performed. This is due to
the fact that many motions correspond to the given acceleration a 5 f(t).
In order to define the motion of the particle uniquely, it is necessary to
specify the initial conditions of the motion, i.e., the value v0 of the
velocity and the value x0 of the position coordinate at t 5 0. Rather
than use an arbitrary constant that is determined by the initial conditions,
it is often more convenient to replace the indefinite integrals with
definite integrals. Definite integrals have lower limits corresponding to
the initial conditions t 5 0 and v 5 v0 and upper limits corresponding
to t 5 t and v 5 v. This gives us
#
v
t
dv 5
# f(t) dt
0
v0
t
v 2 v0 5
# f(t) dt
0
which yields v in terms of t.
We can now solve Eq. (11.1) for dx as
dx 5 v dt
and substitute for v the expression obtained from the first integration.
Then we integrate both sides of this equation via the left-hand side with
respect to x from x 5 x0 to x 5 x and the right-hand side with respect
to t from t 5 0 to t 5 t. In this way, we obtain the position coordinate
x in terms of t; the motion is completely determined.
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Kinematics of Particles
We will study two important cases in greater detail in Sec. 11.2: the
case when a 5 0, corresponding to a uniform motion, and the case when
a 5 constant, corresponding to a uniformly accelerated motion.
2. a 5 f(x). The Acceleration Is a Given Function of x. Rearranging
Eq. (11.4) and substituting f(x) for a, we have
v dv 5 a dx
v dv 5 f(x) dx
Since each side contains only one variable, we can integrate the equation.
Denoting again the initial values of the velocity and of the position
coordinate by v0 and x0, respectively, we obtain
v
x
v0
x0
# v dv 5 # f(x) dx
x
1 2
2v
2 12 v20 5
# f(x) dx
x0
which yields v in terms of x. We now solve Eq. (11.1) for dt, giving
dt 5
dx
v
and substitute for v the expression just obtained. We can then integrate
both sides to obtain the desired relation between x and t. However, in
most cases, this last integration cannot be performed analytically, and
we must resort to a numerical method of integration.
3. a 5 f(v). The Acceleration Is a Given Function of v. We can now
substitute f(v) for a in either Eqs. (11.2) or (11.4) to obtain either
f(v) 5
dt 5
dv
dt
f(v) 5 v
dv
f(v)
dx 5
dv
dx
v dv
f(v)
Integration of the first equation yields a relation between v and t;
integration of the second equation yields a relation between v and x.
Either of these relations can be used in conjunction with Eq. (11.1) to
obtain the relation between x and t that characterizes the motion of the
particle.
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11.1
623
Rectilinear Motion of Particles
Sample Problem 11.1
The position of a particle moving along a straight line is defined by the
relation x 5 t3 2 6t2 2 15t 1 40, where x is expressed in meters and t
in seconds. Determine (a) the time at which the velocity is zero, (b) the
position and distance traveled by the particle at that time, (c) the acceleration of the particle at that time, (d) the distance traveled by the particle
from t 5 4 s to t 5 6 s.
STRATEGY: You need to use the basic kinematic relationships between
position, velocity, and acceleration. Because the position is given as a
function of time, you can differentiate it to find equations for the velocity
and acceleration. Once you have these equations, you can solve the problem.
MODELING and ANALYSIS: Taking the derivative of position, you obtain
x 5 t3 2 6t2 2 15t 1 40
(1)
x (m)
dx
5 3t 2 2 12t 2 15
dt
dv
a5
5 6t 2 12
dt
(2)
v5
40
(3)
+5
0
t (s)
These equations are graphed in Fig. 1.
a. Time at Which v 5 0.
Set v 5 0 in Eq. (2) for
2
3t 2 12t 2 15 5 0
– 60
t 5 21 s
t 5 15 s b
and
Only the root t 5 15 s corresponds to a time after the motion has begun:
for t , 5 s, v , 0 and the particle moves in the negative direction; for
t . 5 s, v . 0 and the particle moves in the positive direction.
v (m/s)
b. Position and Distance Traveled When v 5 0.
Substitute
t 5 15 s into Eq. (1), yielding
+5
0
x5 5 (5)3 2 6(5)2 2 15(5) 1 40
t (s)
x5 5 260 m b
The initial position at t 5 0 was x0 5 140 m. Since v Þ 0 during the
interval t 5 0 to t 5 5 s, you have
Distance traveled 5 x5 2 x0 5 260 m 2 40 m 5 2100 m
Distance traveled 5 100 m in the negative direction b
a (m/s 2)
c. Acceleration When v 5 0.
a5 5 6(5) 2 12
18
0
Substitute t 5 15 s into Eq. (3) for
+2
+5
t (s)
Fig. 1 Motion curves for the particle.
a5 5 118 m/s2 b
d. Distance Traveled from t 5 4 s to t 5 6 s. The particle
moves in the negative direction from t 5 4 s to t 5 5 s and in the positive
direction from t 5 5 s to t 5 6 s; therefore, the distance traveled during
each of these time intervals must be computed separately.
From t 5 4 s to t 5 5 s:
x5 5 260 m
x4 5 (4)3 2 6(4)2 2 15(4) 1 40 5 252 m
(continued)
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Kinematics of Particles
Distance traveled 5 x5 2 x4 5 260 m 2 (252 m) 5 28 m
5 8 m in the negative direction
From t 5 5 s to t 5 6 s:
x5 5 260 m
x6 5 (6)3 2 6(6)2 2 15(6) 1 40 5 250 m
Distance traveled 5 x6 2 x5 5 250 m 2 (260 m) 5 110 m
5 10 m in the positive direction
Total distance traveled from t 5 4 s to t 5 6 s is 8 m 1 10 m 5 18 m
REFLECT and THINK: The total distance traveled by the particle in
the 2-second interval is 18 m, but because one distance is positive and
one is negative, the net change in position is only 2 m (in the positive
direction). This illustrates the difference between total distance traveled
and net change in position. Note that the maximum displacement occurs at
t 5 5 s, when the velocity is zero.
Sample Problem 11.2
You throw a ball vertically upward with a velocity of 10 m/s from a
window located 20 m above the ground. Knowing that the acceleration of
the ball is constant and equal to 9.81 m/s2 downward, determine (a) the
velocity v and elevation y of the ball above the ground at any time t,
(b) the highest elevation reached by the ball and the corresponding value
of t, (c) the time when the ball hits the ground and the corresponding
velocity. Draw the v−t and y−t curves.
y
v0 = +10 m/s
a = – 9.81 m/s2
y0 = +20 m
O
Fig. 1 Acceleration, initial
v (m /s)velocity, and initial position of
the ball.Velocity-time curve
10
Sl
op
e
=
0
1.019 a =
3.28 t (s)
–9
.8
1
m
/s 2
STRATEGY: The acceleration is constant, so you can integrate the
defining kinematic equation for acceleration once to find the velocity
equation and a second time to find the position relationship. Once you
have these equations, you can solve the problem.
MODELING and ANALYSIS: Model the ball as a particle with
negligible drag.
a. Velocity and Elevation. Choose the y axis measuring the position
coordinate (or elevation) with its origin O on the ground and its positive
sense upward. The value of the acceleration and the initial values of v
and y are as indicated in Fig. 1. Substituting for a in a 5 dv/dt and noting
that, when t 5 0, v0 5 110 m/s, you have
dv
5 a 5 29.81 m/s2
dt
#
v
dv 5 2
v0510
#
t
9.81 dt
0
[v] v10 5 2[9.81t] t0
v 2 10 5 29.81t
v 5 10 2 9.81t
–22.2
pe
Slo
e=
25.1
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Slop
=
v0
=
10
m
/s
y (m)
(1) b
12/10/15 11:36 AM
va0== –+10
9.81m/s
m/s2
a =m– 9.81 m/s2
y0 = +20
O
11.1
y0 = +20 m
O
v (m /s)
Velocity-time curve
10
Sl
op
v (m /s)
e
=
0
a
curve
3.28 t (s)
=
10 1.019 Velocity-time
–9
Sl
op
.8
1
e
m
=
/s 2
0
a
1.019
3.28 t (s)
=
–9
.8
1
m
–22.2
/s 2
Fig.
2 Velocity of the ball as a
–22.2
y (m)
function
of time.
/s
10
m
=
/s
m
10
=
Slo
v0
=
pe
Slo
Position-time
curve
Position-time
0
curve
1.019
0
#
dt
y
5 v 5 10 2 9.81t
t
dy 5
y 0520
# (10 2 9.81t) dt
0
[y ] y20 5 [10t 2 4.905t 2 ] t0
y 2 20 5 10t 2 4.905t 2
y 5 20 1 10t 2 4.905t2 (2) b
b. Highest Elevation. The ball reaches its highest elevation when
v 5 0. Substituting into Eq. (1), you obtain
1.019
t 5 1.019 s b
10 2 9.81t 5 0
/s
/s
.2 m
2m
– 22 = – 22.
v=
v
e=
Slop
20
dy
Graphs of these equations are shown in Figs. 2 and 3.
e=
20
25.1
Substituting for v in v 5 dy/dt and noting that when t 5 0, y0 5 20 m,
you have
Slop
25.1
pe
=
v0
y (m)
625
Rectilinear Motion of Particles
Substituting t 5 1.019 s into Eq. (2), you find
y 5 20 1 10(1.019) 2 4.905(1.019)2
3.28
t (s)
3.28
t (s)
Fig. 3 Height of the ball as a
function of time.
y 5 25.1 m b
c. Ball Hits the Ground. The ball hits the ground when y 5 0.
Substituting into Eq. (2), you obtain
20 1 10t 2 4.905t2 5 0
t 5 21.243 s
and
t 5 13.28 s b
Only the root t 5 13.28 s corresponds to a time after the motion has
begun. Carrying this value of t into Eq. (1), you find
v 5 10 2 9.81(3.28) 5 222.2 m/s
v 5 22.2 m/s w b
REFLECT and THINK: When the acceleration is constant, the velocity
changes linearly, and the position is a quadratic function of time. You will
see in Sec. 11.2 that the motion in this problem is an example of free fall,
where the acceleration in the vertical direction is constant and equal to 2g.
Piston
x
Oil
Sample Problem 11.3
Many mountain bike shocks utilize a piston that travels in an oil-filled
cylinder to provide shock absorption; this system is shown schematically.
When the front tire goes over a bump, the cylinder is given an initial
velocity v0. The piston, which is attached to the fork, then moves with
respect to the cylinder, and oil is forced through orifices in the piston.
This causes the piston to decelerate at a rate proportional to the velocity
at a 5 2kv. At time t 5 0, the position of the piston is x 5 0. Express
(a) the velocity v in terms of t, (b) the position x in terms of t, (c) the
velocity v in terms of x. Draw the corresponding motion curves.
(continued)
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626
Kinematics of Particles
STRATEGY: Because the acceleration is given as a function of velocity,
you need to use either a 5 dv/dt or a 5 v dv/dx and then separate variables
and integrate. Which one you use depends on what you are asked to find.
Since part a asks for v in terms of t, use a 5 dv/dt. You can integrate this
again using v 5 dx/dt for part b. Since part c asked for v(x), you should
use a 5 v dv/dx and then separate the variables and integrate.
MODELING and ANALYSIS: Rotation of the piston is not relevant,
so you can model it as a particle undergoing rectilinear motion.
a. v in Terms of t. Substitute 2kv for a in the fundamental formula
defining acceleration, a 5 dv/dt. You obtain
2kv 5
dv
dt
#
dv
5 2k dt
v
ln
v
v0
dv
5 2k
v
#
v
5 2kt
v0
t
dt
0
v 5 v0 e2kt b
b. x in Terms of t. Substitute the expression just obtained for v into
v 5 dx/dt. You get
v0 e2kt 5
v
v0
#
dx
dt
x
t
dx 5 v0
0
x52
O
#e
2kt
v0 2kt t
v0
[e ] 0 5 2 (e2kt 2 1)
k
k
t
x
dt
0
x5
v0
(1 2 e2kt )
k
b
c. v in Terms of x. Substitute 2kv for a in a 5 v dv/dx. You have
v0
k
2kv 5 v
dv
dx
dv 5 2k dx
v
O
x
# dv 5 2k # dx
t
v0
v
v0
0
v 2 v0 5 2kx
v 5 v0 2 kx
b
The motion curves are shown in Fig. 1.
O
v0
x
REFLECT and THINK: You could have solved part c by eliminating t
from the answers obtained for parts a and b. You could use this alternative
method as a check. From part a, you obtain e2kt 5 v/v0; substituting into
the answer of part b, you have
k
Fig. 1 Motion curves for the
piston
bee87342_ch11_615-717.indd 626
x5
v0
v0
v
(1 2 e2kt ) 5 a1 2 b
v0
k
k
v 5 v0 2 kx
(checks)
12/10/15 11:36 AM
11.1
627
Rectilinear Motion of Particles
Sample Problem 11.4
An uncontrolled automobile traveling at 72 km/h strikes a highway crash
barrier square on. After initially hitting the barrier, the automobile decelerates at a rate proportional to the distance x the automobile has moved into
the barrier; specifically, a 5 2302x, where a and x are expressed in m/s2
and m, respectively. Determine the distance the automobile will move into
the barrier before it comes to rest.
y
–a (m/s2)
v0
x
z
x (m)
STRATEGY: Since you are given the deceleration as a function of
displacement, you should start with the basic kinematic relationship
a 5 v dv/dx.
MODELING and ANALYSIS: Model the car as a particle. First find
the initial speed in ft/s,
v0 5 a72
km
1 hr
1000 m
m
ba
ba
b 5 20
s
hr 3600 s
km
Substituting a 5 2302x into a 5 v dv/dx gives
a 5 2302x 5
v dv
dx
Separating variables and integrating gives
v dv 5 2302x dx y
#
0
x
v dv 5 2
# 302x dx
0
v0
2/3
1 2 1 2
1
v 2 v0 5 220x3/2 y x 5 a (v20 2 v2 )b
2
2
40
(1)
Substituting v 5 0, v0 5 20 m/s gives
d 5 4.64 m
b
REFLECT and THINK: A distance of 4.64 m seems reasonable for a
barrier of this type. If you substitute d into the equation for a, you find a
maximum deceleration of about 7 g’s. Note that this problem would have
been much harder to solve if you had been asked to find the time for the
automobile to stop. In this case, you would need to determine v(t) from
Eq. (1). This gives v 5 2v20 2 40x3/2. Using the basic kinematic relationship v 5 dx/dt, you can easily show that
#
t
0
x
dt 5
# 2v
0
dx
2
0
2 40x3/2
Unfortunately, there is no closed-form solution to this integral, so you
would need to solve it numerically.
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SOLVING PROBLEMS
ON YOUR OWN
I
n the problems for this section, you will be asked to determine the position,
velocity, and/or acceleration of a particle in rectilinear motion. As you read each
problem, it is important to identify both the independent variable (typically t or x)
and what is required (for example, the need to express v as a function of x). You may
find it helpful to start each problem by writing down both the given information and
a simple statement of what is to be determined.
1. Determining v(t) and a(t) for a given x(t). As explained in Sec. 11.1A, the first
and second derivatives of x with respect to t are equal to the velocity and the acceleration, respectively, of the particle [Eqs. (11.1) and (11.2)]. If the velocity and acceleration have opposite signs, the particle can come to rest and then move in the opposite
direction [Sample Prob. 11.1]. Thus, when computing the total distance traveled by a
particle, you should first determine if the particle comes to rest during the specified
interval of time. Constructing a diagram similar to that of Sample Prob. 11.1, which
shows the position and the velocity of the particle at each critical instant (v 5 vmax,
v 5 0, etc.), will help you to visualize the motion.
2. Determining v(t) and x(t) for a given a(t). We discussed the solution of problems
of this type in the first part of Sec. 11.1B. We used the initial conditions, t 5 0 and
v 5 v0, for the lower limits of the integrals in t and v, but any other known state (for
example, t 5 t1 and v 5 v1) could be used instead. Also, if the given function a(t)
contains an unknown constant (for example, the constant k if a 5 kt), you will first
have to determine that constant by substituting a set of known values of t and a in
the equation defining a(t).
3. Determining v(x) and x(t) for a given a(x). This is the second case considered
in Sec. 11.1B. We again note that the lower limits of integration can be any known
state (for example, x 5 x1 and v 5 v1). In addition, since v 5 vmax when a 5 0, you
can determine the positions where the maximum or minimum values of the velocity
occur by setting a(x) 5 0 and solving for x.
4. Determining v(x), v(t), and x(t) for a given a(v). This is the last case treated in
Sec. 11.1B; the appropriate solution techniques for problems of this type are illustrated
in Sample Probs. 11.3 and 11.4. All of the general comments for the preceding cases
once again apply. Note that Sample Prob. 11.3 provides a summary of how and when
to use the equations v 5 dx/dt, a 5 dv/dt, and a 5 v dv/dx.
628
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We can summarize these relationships in Table 11.1.
Table 11.1
If....
Kinematic relationship
a 5 a(t)
dv
5 a(t)
dt
a 5 a(x)
v
dv
5 a(x)
dx
t
v0
0
#
v
x
v dv 5
v0
# a(x)dx
x0
v
t
#
dv
5 a(v)
dx
# dx 5 # a(v)
v
bee87342_ch11_615-717.indd 629
v
# dv 5 # a(t)dt
dv
5 a(v)
dt
a 5 a(v)
629
Integrate
v0
dv
5
a(v)
x
v
x0
v0
# dt
0
v dv
629
12/10/15 11:36 AM
Problems†
CONCEPT QUESTIONS
B
11.CQ1 A bus travels the 100 km between A and B at 50 km/h and then
another 100 km between B and C at 70 km/h. The average speed
of the bus for the entire 200 km trip is:
a. More than 60 km/h.
b. Equal to 60 km/h.
c. Less than 60 km/h.
A
C
Fig. P11.CQ1
11.CQ2 Two cars A and B race each other down a straight road. The posi-
tion of each car as a function of time is shown. Which of the following statements are true (more than one answer can be correct)?
a. At time t2 both cars have traveled the same distance.
b. At time t1 both cars have the same speed.
c. Both cars have the same speed at some time t , t1.
d. Both cars have the same acceleration at some time t , t1.
e. Both cars have the same acceleration at some time t1 , t , t2.
Position
B
A
t1
t2 time
Fig. P11.CQ2
END-OF-SECTION PROBLEMS
11.1 A snowboarder starts from rest at the top of a double black diamond
hill. As she rides down the slope, GPS coordinates are used to determine her displacement as a function of time: x 5 0.5t3 1 t2 1 2t,
where x and t are expressed in meters and seconds, respectively.
Determine the position, velocity, and acceleration of the boarder
when t 5 5 seconds.
11.2 The motion of a particle is defined by the relation x 5 2t3 2 9t2 1
12t 1 10, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the acceleration of the
particle when v 5 0.
11.3 The vertical motion of mass A is defined by the relation x 5
10 sin 2t 1 15 cos 2t 1 100, where x and t are expressed in
millimeters and seconds, respectively. Determine (a) the position,
velocity, and acceleration of A when t 5 1 s, (b) the maximum
velocity and acceleration of A.
A
Fig. P11.3
†
Answers to all problems set in straight type (such as 11.1) are given at the end of the book.
Answers to problems with a number set in italic type (such as 11.6) are not given.
630
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11.4 A loaded railroad car is rolling at a constant velocity when it couples
with a spring and dashpot bumper system. After the coupling, the
motion of the car is defined by the relation x 5 60e24.8t sin 16t, where
x and t are expressed in millimeters and seconds, respectively. Determine the position, the velocity, and the acceleration of the railroad
car when (a) t 5 0, (b) t 5 0.3 s.
v0
k
c
Fig. P11.4
11.5 The motion of a particle is defined by the relation x 5 6t4 2 2t3 2 12t2 1
3t 1 3, where x and t are expressed in meters and seconds,
respectively. Determine the time, the position, and the velocity when
a 5 0.
11.6 The motion of a particle is defined by the relation x 5 t3 2 9t2 1
24t 2 8, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and
the total distance traveled when the acceleration is zero.
11.7 A girl operates a radio-controlled model car in a vacant parking lot.
The girl’s position is at the origin of the xy coordinate axes, and the
surface of the parking lot lies in the x-y plane. She drives the car in a
straight line so that the x coordinate is defined by the relation
x(t) 5 0.5t3 2 3t2 1 3t 1 2, where x and t are expressed in meters
and seconds, respectively. Determine (a) when the velocity is zero,
(b) the position and total distance travelled when the acceleration is zero.
y (m)
6
0
2
x (m)
Fig. P11.7
11.8 The motion of a particle is defined by the relation x 5 t 2 2 (t 2 2) 3,
where x and t are expressed in meters and seconds, respectively.
Determine (a) the two positions at which the velocity is zero (b) the
total distance traveled by the particle from t 5 0 to t 5 4 s.
631
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v0
v=0
A
100 m
11.9 The brakes of a car are applied, causing it to slow down at a rate of
3 m/s2. Knowing that the car stops in 100 m, determine (a) how fast
the car was traveling immediately before the brakes were applied,
(b) the time required for the car to stop.
11.10 The acceleration of a particle is defined by the relation a 5 3e20.2t,
where a and t are expressed in m/s2 and seconds, respectively.
Knowing that x 5 0 and v 5 0 at t 5 0, determine the velocity and
position of the particle when t 5 0.5 s.
x
Fig. P11.9
11.11 The acceleration of a particle is directly proportional to the square of
the time t. When t 5 0, the particle is at x 5 24 m. Knowing that at
t 5 6 s, x 5 96 m and v 5 18 m/s, express x and v in terms of t.
11.12 The acceleration of a particle is defined by the relation a 5 kt2.
(a) Knowing that v 5 28 m/s when t 5 0 and that v 5 18 m/s when
t 5 2 s, determine the constant k. (b) Write the equations of motion,
knowing also that x 5 0 when t 5 2 s.
x
A
C
B
D
Fig. P11.13 and P11.14
11.13 A Scotch yoke is a mechanism that transforms the circular motion
of a crank into the reciprocating motion of a shaft (or vice versa).
It has been used in a number of different internal combustion engines
and in control valves. In the Scotch yoke shown, the acceleration of
point A is defined by the relation a 5 21.8 sin kt, where a and t
are expressed in m/s2 and seconds, respectively, and k 5 3 rad/s.
Knowing that x 5 0 and v 5 0.6 m/s when t 5 0, determine the
velocity and position of point A when t 5 0.5 s.
11.14 For the Scotch yoke mechanism shown, the acceleration of point A
is defined by the relation a 521.08 sin kt 2 1.44 cos kt, where a
and t are expressed in m/s2 and seconds, respectively, and
k 5 3 rad/s. Knowing that x 5 0.16 m and v 5 0.36 m/s when
t 5 0, determine the velocity and position of point A when t 5 0.5 s.
11.15 A piece of electronic equipment that is surrounded by packing material
is dropped so that it hits the ground with a speed of 4 m/s. After
contact the equipment experiences an acceleration of a 5 2kx, where
k is a constant and x is the compression of the packing material. If the
packing material experiences a maximum compression of 20 mm,
determine the maximum acceleration of the equipment.
v
x
Fig. P11.15
v
Fig. P11.16
11.16 A projectile enters a resisting medium at x 5 0 with an initial velocity
v0 5 270 m/s and travels 100 mm. before coming to rest.
Assuming that the velocity of the projectile is defined by the relation
v 5 v0 2 kx, where v is expressed in m/s and x is in meters, determine
(a) the initial acceleration of the projectile, (b) the time required for
the projectile to penetrate 97.5 mm into the resisting medium.
632
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11.17 The acceleration of a particle is defined by the relation
It has been experimentally determined that v 5 5
x 5 0.2 m and that v 5 3 m/s when x 5 0.4 m.
(a) the velocity of the particle when x 5 0.5 m, (b) the
the particle at which its velocity is zero.
a 5 2k/x.
m/s when
Determine
position of
A
B
11.18 A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass tube under the magnetic repelling force of another
steel magnet C located at a distance x 5 0.004 m from B. The force
is inversely proportional to the square of the distance between B and
C. If block A is suddenly removed, the acceleration of block B is
a 5 29.81 1 k/x2, where a and x are expressed in m/s2 and meters,
respectively, and k 5 4 3 1024 m3/s2. Determine the maximum
velocity and acceleration of B.
11.19 Based on experimental observations, the acceleration of a particle is
defined by the relation a 5 2(0.1 1 sin x/b), where a and x are
expressed in m/s2 and meters, respectively. Knowing that b 5 0.8 m
and that v 5 1 m/s when x 5 0, determine (a) the velocity of the
particle when x 5 21 m, (b) the position where the velocity is
maximum, (c) the maximum velocity.
11.20 A spring AB is attached to a support at A and to a collar. The
unstretched length of the spring is l. Knowing that the collar is
released from rest at x 5 x0 and has an acceleration defined by the
relation a 5 2100(x 2 lx/ 2l2 1 x2 ), determine the velocity of the
collar as it passes through point C.
x
C
Fig. P11.18
A
l
B
C
x0
Fig. P11.20
11.21 The acceleration of a particle is defined by the relation a 5 k(1 2 e2x ),
where k is a constant. Knowing that the velocity of the particle is
v 5 19 m/s when x 5 23 m and that the particle comes to rest at
the origin, determine (a) the value of k, (b) the velocity of the
particle when x 5 22 m.
11.22 Starting from x 5 0 with no initial velocity, a particle is given an
acceleration a 5 0.12v2 1 49, where a and v are expressed in m/s2
and m/s, respectively. Determine (a) the position of the particle when
v 5 24 m/s, (b) the speed and acceleration of the particle when
x 5 40 m.
11.23 A bowling ball is dropped from a boat so that it strikes the surface of
a lake with a speed of 8 m/s. Assuming the ball experiences a downward acceleration of a 5 3 2 0.1v2 (where a and v are expressed in
m/s2 and m/s, respectively) when in the water, determine the velocity
of the ball when it strikes the bottom of the lake.
10 m
11.24 The acceleration of a particle is defined by the relation a 5 2k 1v,
where k is a constant. Knowing that x 5 0 and v 5 81 m/s at
t 5 0 and that v 5 36 m/s when x 5 18 m, determine (a) the velocity
of the particle when x 5 20 m, (b) the time required for the particle
to come to rest.
11.25 The acceleration of a particle is defined by the relation a 5 2kv2.5,
where k is a constant. The particle starts at x 5 0 with a velocity of
16 mm/s, and when x 5 6 mm, the velocity is observed to be 4 mm/s.
Determine (a) the velocity of the particle when x 5 5 mm, (b) the
time at which the velocity of the particle is 9 mm/s.
Fig. P11.23
633
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12/10/15 11:37 AM
11.26 A human-powered vehicle (HPV) team wants to model the acceleration during the 260-m sprint race (the first 60 m is called a flying
start) using a 5 A 2 Cv2, where a is acceleration in m/s2 and v is
the velocity in m/s. From wind tunnel testing, they found that
C 5 0.0012 m21. Knowing that the cyclist is going 100 km/h at the
260-meter mark, what is the value of A?
11.27 Experimental data indicate that in a region downstream of a given
louvered supply vent the velocity of the emitted air is defined by
v 5 0.18v0 /x, where v and x are expressed in m/s and meters,
respectively, and v0 is the initial discharge velocity of the air. For
v0 5 3.6 m/s, determine (a) the acceleration of the air at x 5 2 m,
(b) the time required for the air to flow from x 5 1 to x 5 3 m.
11.28 Based on observations, the speed of a jogger can be approximated by
the relation v 5 12(1 2 0.06x)0.3, where v and x are expressed in
km/h and km, respectively. Knowing that x 5 0 at t 5 0, determine
(a) the distance the jogger has run when t 5 1 h, (b) the jogger’s
acceleration in m/s2 at t 5 0, (c) the time required for the jogger to
run 9 Km.
11.29 The acceleration due to gravity at an altitude y above the surface of
the earth can be expressed as
29.81
a5
Fig. P11.26
v
11
x
Fig. P11.27
v
Fig. P11.28
P
P
y
r
R
where a and y are expressed in m/s2 and metre, respectively. Using
this expression, compute the height reached by a projectile fired
vertically upward from the surface of the earth if its initial velocity
is (a) 540 m/s, (b) 900 m/s, (c) 11,180 m/s.
11.30 The acceleration due to gravity of a particle falling toward the
earth is a 5 2gR2/r2, where r is the distance from the center of
the earth to the particle, R is the radius of the earth, and g is the
acceleration due to gravity at the surface of the earth. If
R 5 6370 km, calculate the escape velocity, that is, the minimum
velocity with which a particle must be projected vertically upward
from the surface of the earth if it is not to return to the earth.
(Hint: v 5 0 for r 5 `.)
11.31 The velocity of a particle is v 5 v0[1 2 sin(πt/T)]. Knowing that
the particle starts from the origin with an initial velocity v0, determine (a) its position and its acceleration at t 5 3T, (b) its average
velocity during the interval t 5 0 to t 5 T.
11.32 An eccentric circular cam, which serves a similar function as the
Scotch yoke mechanism in Problem 11.13, is used in conjunction
with a flat face follower to control motion in pumps and in steam
engine valves. Knowing that the eccentricity is denoted by e, the
maximum range of the displacement of the follower is dmax and the
maximum velocity of the follower is vmax, determine the displacement, velocity, and acceleration of the follower.
e
Fig. P11.29
Fig. P11.30
O
θ
r
A
y
Fig. P11.32
634
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11.2
11.2
Special Cases and Relative Motion
635
SPECIAL CASES AND
RELATIVE MOTION
In this section, we derive the equations that describe uniform rectilinear
motion and uniformly accelerated rectilinear motion. We also introduce
the concept of relative motion, which is of fundamental importance whenever we consider the motion of more than one particle at the same time.
11.2A
Uniform Rectilinear Motion
Uniform rectilinear motion is a type of straight-line motion that is frequently encountered in practical applications. In this motion, the acceleration a of the particle is zero for every value of t. The velocity v is therefore
constant, and Eq. (11.1) becomes
dx
5 v 5 constant
dt
We can obtain the position coordinate x by integrating this equation.
Denoting the initial value of x by x0, we have
Distance in uniform
rectilinear motion
#
x
t
dx 5 v
x0
# dt
0
x 2 x0 5 vt
x 5 x0 1 vt
(11.5)
This equation can be used only if the velocity of the particle is known to
be constant. For example, this would be true for an airplane in steady
flight or a car cruising along a highway at a constant speed.
11.2B
Uniformly Accelerated
Rectilinear Motion
Uniformly accelerated rectilinear motion is another common type of
motion. In this case, the acceleration a of the particle is constant, and
Eq. (11.2) becomes
dv
5 a 5 constant
dt
We obtain the velocity v of the particle by integrating this equation as
v
t
v0
0
# dv 5 a # dt
v 2 v0 5 at
v 5 v0 1 at
(11.6)
where v0 is the initial velocity. Substituting for v in Eq, (11.1), we have
dx
5 v0 1 at
dt
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636
Kinematics of Particles
Denoting by x0 the initial value of x and integrating, we have
#
x
t
dx 5
# (v
0
x0
0
1 at)dt
x 2 x0 5 v0 t 1 12 at 2
x 5 x0 1 v0 t 1 12 at 2
(11.7)
We can also use Eq. (11.4) and write
dv
5 a 5 constant
dx
v dv 5 a dx
v
Integrating both sides, we obtain
#
v
x
v dv 5 a
v0
1
2
2 (v
# dx
x0
2 v20 ) 5 a(x 2 x0 )
v 2 5 v20 1 2a(x 2 x0 )
(11.8)
The three equations we have derived provide useful relations among
position, velocity, and time in the case of constant acceleration, once you have
provided appropriate values for a, v0, and x0. You first need to define the origin
O of the x axis and choose a positive direction along the axis; this direction
determines the signs of a, v0, and x0. Equation (11.6) relates v and t and should
be used when the value of v corresponding to a given value of t is desired,
or inversely. Equation (11.7) relates x and t; Eq. (11.8) relates v and x. An
important application of uniformly accelerated motion is the motion of a body
in free fall. The acceleration of a body in free fall (usually denoted by g) is
equal to 9.81 m/s2 or 32.2 ft/s2 (we ignore air resistance in this case).
It is important to keep in mind that the three equations can be used
only when the acceleration of the particle is known to be constant. If the
acceleration of the particle is variable, you need to determine its motion
from the fundamental Eqs. (11.1) through (11.4) according to the methods
outlined in Sec. 11.1B.
11.2C
A
O
When several particles move independently along the same line, you can
write independent equations of motion for each particle. Whenever
possible, you should record time from the same initial instant for all
particles and measure displacements from the same origin and in the same
direction. In other words, use a single clock and a single measuring tape.
B
xB/A
xA
x
xB
Fig. 11.7 Two particles A and B in motion
along the same straight line.
Motion of Several Particles
Relative Motion of Two Particles. Consider two particles A and
B moving along the same straight line (Fig. 11.7). If we measure the
position coordinates xA and xB from the same origin, the difference xB 2 xA
defines the relative position coordinate of B with respect to A, which
is denoted by xB/A. We have
Relative position
of two particles
xB/A 5 xB 2 xA
or
xB 5 xA 1 xB/A
(11.9)
Regardless of the positions of A and B with respect to the origin, a positive
sign for xB/A means that B is to the right of A, and a negative sign means
that B is to the left of A.
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11.2
Special Cases and Relative Motion
637
The rate of change of xB/A is known as the relative velocity of B with
respect to A and is denoted by vB/A. Differentiating Eq. (11.9), we obtain
Relative velocity
of two particles
vB/A 5 vB 2 vA
or
vB 5 vA 1 vB/A
(11.10)
A positive sign for vB/A means that B is observed from A to move in the
positive direction; a negative sign means that it is observed to move in
the negative direction.
The rate of change of vB/A is known as the relative acceleration of B
with respect to A and is denoted by aB/A. Differentiating Eq. (11.10), we obtain†
Relative acceleration
of two particles
aB/A 5 aB 2 aA
or
aB 5 aA 1 aB/
B/A
B
/A
/A
(11.11)
Dependent Motion of Particles. Sometimes, the position of a
particle depends upon the position of another particle or of several other
particles. These motions are called dependent. For example, the position
of block B in Fig. 11.8 depends upon the position of block A. Since the
rope ACDEFG is of constant length, and since the lengths of the portions
of rope CD and EF wrapped around the pulleys remain constant, it follows
that the sum of the lengths of the segments AC, DE, and FG is constant.
Observing that the length of the segment AC differs from xA only by a
constant and that, similarly, the lengths of the segments DE and FG differ
from xB only by a constant, we have
xA 1 2xB 5 constant
Since only one of the two coordinates xA and xB can be chosen arbitrarily, we
say that the system shown in Fig. 11.8 has one degree of freedom. From the
relation between the position coordinates xA and xB, it follows that if xA is
given an increment DxA––that is, if block A is lowered by an amount DxA––the
coordinate xB receives an increment DxB 5 212 DxA. In other words, block B
rises by half the same amount. You can check this directly from Fig. 11.8.
In the case of the three blocks of Fig. 11.9, we can again observe
that the length of the rope that passes over the pulleys is constant.
Thus, the following relation must be satisfied by the position coordinates
of the three blocks:
Photo 11.2 Multiple cables and pulleys are
used by this shipyard crane.
C
G
D
xA
xB
A
E
F
B
Fig. 11.8 A system of blocks and pulleys
with one degree of freedom.
2xA 1 2xB 1 xC 5 constant
Since two of the coordinates can be chosen arbitrarily, we say that the
system shown in Fig. 11.9 has two degrees of freedom.
When the relation existing between the position coordinates of several
particles is linear, a similar relation holds between the velocities and between
the accelerations of the particles. In the case of the blocks of Fig. 11.9, for
instance, we can differentiate the position equation twice and obtain
dxC
dxA
dxB
12
1
50
dt
dt
dt
dvC
dvA
dvB
12
1
50
2
dt
dt
dt
2
†
C
xB
A
or
2vA 1 2vB 1 vC 5 0
or
2aA 1 2aB 1 aC 5 0
Note that the product of the subscripts A and B/A used in the right-hand sides of Eqs. (11.9),
(11.10), and (11.11) is equal to the subscript B that appears in the left-hand sides. This may
help you remember the correct order of subscripts in various situations.
bee87342_ch11_615-717.indd 637
xC
xA
B
Fig. 11.9 A system of blocks and pulleys
with two degrees of freedom.
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638
Kinematics of Particles
Sample Problem 11.5
In an elevator shaft, a ball is thrown vertically upward with an initial
velocity of 18 m/s from a height of 12 m above ground. At the same
instant, an open-platform elevator passes the 5-m level, moving upward
with a constant velocity of 2 m/s. Determine (a) when and where the ball
hits the elevator (b) the relative velocity of the ball with respect to the
elevator when the ball hits the elevator.
t=t
t=t
v0 = 18 m/s
t = 0v0 = 18 m/s
a = –9.81 m/s2
yB
t=0
t=t
a = –9.81 m/s2
vy0 == 18
12 m/s
m
yB
0
O
t=0
y0 = 12 m
a = –9.81 m/s2
yB 1 Acceleration,
O
Fig.
initial
velocity, and initial position of
the ball.
y0 = 12 m
O
yE
t=0
2 m/s
y v=E5=m
t =0t
t=0
y0 = 5 m
O
yE
Motion of Ball. Place the origin O of the y axis at ground level and
choose its positive direction upward (Fig. 1). Then the initial position of
the ball is y0 5 112 m, its initial velocity is v0 5 118 m/s, and its acceleration is a 5 29.81 m/s2. Substituting these values in the equations for
uniformly accelerated motion, you get
vB 5 v0 1 at
yB 5 y0 1 v0 t 1
vE = 2 m/s
O
MODELING and ANALYSIS:
t=t
t=t
yE
STRATEGY: The ball has a constant acceleration, so its motion is uniformly accelerated. The elevator has a constant velocity, so its motion is
uniform. You can write equations to describe each motion and then set the
position coordinates equal to each other to find when the particles meet. The
relative velocity is determined from the calculated motion of each particle.
vE = 2 m/s
Fig. 2 Initial velocity and initial
t=0
position of the elevator.
y0 = 5 m
O
yB
yE
yB
yE
1 2
at
2
vB 5 18 2 9.81t
(1)
yB 5 12 1 18t 2 4.905t 2
(2)
Motion of Elevator. Again place the origin O at ground level and
choose the positive direction upward (Fig. 2). Noting that y0 5 15 m,
you have
vE 5 12 m/s
yE 5 y0 1 vE t
(3)
(4)
yE 5 5 1 2t
Ball Hits Elevator. First note that you used the same time t and the
same origin O in writing the equations of motion for both the ball and
the elevator. From Fig. 3, when the ball hits the elevator,
(5)
yE 5 yB
O
O
yB
yE
O
Fig. 3 Position of ball and
elevator at time t.
bee87342_ch11_615-717.indd 638
Substituting for yE and yB from Eqs. (2) and (4) into Eq. (5), you have
5 1 2t 5 12 1 18t 2 4.905t2
t 5 20.39 s
and
t 5 3.65 s
b
Only the root t 5 3.65 s corresponds to a time after the motion has begun.
Substituting this value into Eq. (4), you obtain
yE 5 5 1 2(3.65) 5 12.30 m
Elevation from ground 5 12.30 m b
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11.2
639
Special Cases and Relative Motion
Relative Velocity. The relative velocity of the ball with respect to the
elevator is
vB/E 5 vB 2 vE 5 (18 2 9.81t) 2 2 5 16 2 9.81t
When the ball hits the elevator at time t 5 3.65 s, you have
vB/E 5 16 2 9.81(3.65)
vB/E 5 219.81 m/s b
The negative sign means that if you are riding on the elevator, it will
appear as if the ball is moving downward.
REFLECT and THINK: The key insight is that, when two particles
collide, their position coordinates must be equal. Also, although you can
use the basic kinematic relationships in this problem, you may find it
easier to use the equations relating a, v, x, and t when the acceleration is
constant or zero.
Sample Problem 11.6
Car A is travelling at a constant 135 km/h when she passes a parked police
officer B, who gives chase when the car passes her. The officer accelerates
at a constant rate until she reaches the speed of 150 km/h. Thereafter, her
speed remains constant. The police officer catches the car 4.5 km from
her starting point. Determine the initial acceleration of the police
officer.
STRATEGY: One car is traveling at a constant speed and the other has
a constant acceleration, so you can start with the algebraic relationships
found in Sec. 11.2 rather than separating and integrating the basic kinematic relationships.
MODELING and ANALYSIS: A clearly labeled picture will help you
understand the problem better (Fig. 1). The position, x, is defined from
the point the car passes the officer.
(vA)0 = 135 km/h
aA = 0
(vA)f = 135 km/h
4.5 km
x1
x
t1
(vB)0 = 0
aB = constant
Time when police
officer reaches max
speed
(vB)f = 150 km/h
Fig. 1 Velocities and accelerations of the cars at various times.
(continued)
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640
Kinematics of Particles
Unit Conversions. First you should convert everything to units of feet
and seconds. Use the subscript A for the car and B for the officer
vA 5 a135
1 hr
1000 m
km
m
ba
ba
b 5 37.5
s
hr 3600 s
km
vB 5 a150
1 hr
1000 m
km
125 m
ba
ba
b5
3 s
hr 3600 s
km
Motion of the Speeding Car A. Since the car has a constant speed,
(1)
xA 5 vAt 5 37.5 t
Motion of the Officer B. The officer has a constant acceleration
until she reaches a final speed of 105 mph. This time is labeled t1 in Fig. 1.
Therefore, from time 0 , t , t1, the officer has a velocity of
vB 5 aB t
for 0 , t , t1
or at time t 5 t1, it is
125
5 aBt1
3
(2)
The distance the officer travels is going to be the distance from 0 to t1
and then from t1 to tf. Hence,
xB 5
1 2
aBt1 1 vB (t 2 t1 )
2
for t . t1
(3)
The officer catches the speeder when xA 5 xB 5 4.5 km 5 4,500 m. From
Eq. (1), you can solve for the time tf 5 (4500 m)/(37.5 m/s) 5 120 s.
Therefore, you have two equations: Eq. (2) and
45005
1 2 125
aBt1 1
(120 2 t1 )
2
3
(4)
Substituting Eq. (2) into Eq. (4) allows you to solve for t1:
t1 5 24.0 s
Substituting this into Eq. (2) gives
aB 5 1.736 m/s2 b
REFLECT and THINK: It is important to use the same origin for the
position of both vehicles. The time to accelerate from 0 to 150 km/h seems
reasonable, although it is perhaps longer than you would expect. A highperformance sports car can go from 0 to 90 km/h in less than 5 seconds.
It is very likely that the officer could have accelerated to 150 km/h in less
time if she had wanted to, but perhaps she had to consider the safety of
other motorists.
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11.2
641
Special Cases and Relative Motion
Sample Problem 11.7
C
K
E
Collar A and block B are connected by a cable passing over three pulleys
C, D, and E as shown. Pulleys C and E are fixed, while D is attached to
a collar which is pulled downward with a constant velocity of 75 mm/s. At
t 5 0, collar A starts moving downward from position K with a constant
acceleration and no initial velocity. Knowing that the velocity of collar A
is 300 mm/s as it passes through point L, determine the change in elevation,
the velocity, and the acceleration of block B when collar A passes through
L.
A
D
200 mm
B
L
STRATEGY: You have multiple objects connected by cables, so this is
a problem in dependent motion. Use the given data to write a single equation relating the changes in position coordinates of collar A, pulley D, and
block B. Based on the given information, you will also need to use the
algebraic relationships we found for uniformly accelerated motion.
O
(xA)0
O
xA (xAK
)0
O a
A
xA (xAK)80 in.
xA
MODELING and ANALYSIS:
Motion of Collar A. Place the origin O at the upper horizontal
surface and choose the positive direction downward. Then when t 5 0,
collar A is at position K and (vA)0 5 0 (Fig. 1). Since vA 5 300 mm/s and
xA 2 (xA)0 5 200 mm when the collar passes through L, you have
aA
K8 in.
L
A
8 in.
L
A
aA
vA = 300 mm/s
v2A 5 (vA)20 1 2aA[xA 2 (xA)0]
(300)2 5 0 1 2aA(200)
2
aA 5 225 mm/s
A v = 300 mm/s
A
L O velocity,
Fig. 1 Position,
and
acceleration of collar A.
vA = 300 mm/s
O
(xD)0
O
(xD)0
To find the time at which collar A reaches point L, use the equation for
velocity as a function of time with uniform acceleration. Thus,
vA 5 (vA)0 1 aAt
xD
(xD)0
Motion of Pulley D.
have (Fig. 2)
xD
D
D
O
C
aD 5 0
xD
vD = 75 mm/s
D
E
xB
xA
xA
A
O
BE
C
A
A
D
D
B
Thus,
xD 2 (xD)0 5 100 mm
xD
xA 1 2xD 1 xB 5 (xA)0 1 2(xD)0 1 (xB)0
(1)
[xA 2 (xA)0] 1 2[xD 2 (xD)0] 1 [xB 2 (xB)0] 5 0
(2)
B
D
xD 5 (xD)0 1 vD t 5 (xD)0 1 75t
xD
xB
xB
vD 5 75 mm/s
Motion of Block B. Note that the total length of cable ACDEB differs
from the quantity (xA 1 2xD 1 xB) only by a constant. Since the cable
length is constant during the motion, this quantity must also remain constant. Thus, considering the times t 5 0 and t 5 1.333 s, you can write
xD
E
Since the positive direction is downward, you
xD 5 (xD)0 1 75(1.333) 5 (xD)0 1 100
Fig. 2 PositionOand velocity
ofmm/s
pulley D.
vD = 75
C
But you know that xA 2 (xA)0 5 200 mm and xD 2 (xD)0 5 100 mm
Substituting these values in Eq. (2), you find
200 1 2(100) 1 [xB 2 (xB)0] 5 0
Fig. 3 Position of A, B, and D.
t 5 1.333 s
When collar A reaches L at t 5 1.333 s, the position of pulley D is
vD = 75 mm/s
xA
300 5 0 1 225t
Thus,
xB 2 (xB)0 5 2400 mm
Change in elevation of B 5 400 mmx b
(continued)
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642
Kinematics of Particles
Differentiating Eq. (1) twice, you obtain equations relating the velocities
and the accelerations of A, B, and D. Substituting for the velocities and
accelerations of A and D at t 5 1.333 s, you have
vA 1 2vD 1 vB 5 0:
aA 1 2aD 1 aB 5 0:
300 1 2(75) 1 vB 5 0
vB 5 2450 mm/s
vB 5 450 mm/sx b
225 1 2(0) 1 aB 5 0
aB 5 2225 mm/s2
aB 5 225 mm/s2x b
REFLECT and THINK: In this case, the relationship we needed was
not between position coordinates, but between changes in position
coordinates at two different times. The key step is to clearly define your
position vectors. This is a two-degree-of-freedom system, because two
coordinates are required to completely describe it.
Sample Problem 11.8
Block C starts from rest and moves down with a constant acceleration.
Knowing that after block A has moved 450 mm its velocity is 180 mm/s,
determine (a) the acceleration of A and C, (b) the change in velocity and
the change in position of block B after 2.5 seconds.
B
A
C
STRATEGY: Since you have blocks connected by cables, this is a
dependent-motion problem. You should define coordinates for each mass
and write constraint equations for both cables.
MODELING and ANALYSIS: Define position vectors as shown in
Fig. 1, where positive is defined to be down.
Cable 2
xB
xC
xA
B
C
Cable 1
A
Fig. 1 Position of A, B, and C.
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11.2
643
Special Cases and Relative Motion
Constraint Equations. Assuming the cables are inextensible, you can
write the lengths in terms of the defined coordinates and then
differentiate.
Cable 1:
xA 1 (xA 2 xB ) 5 constant
Differentiating this, you find
2vA 5 vB and
Cable 2:
(1)
2aA 5 aB
2xB 1 xC 5 constant
Differentiating this, you find
vC 5 22vB and
aC 5 22aB
(2)
aC 5 24aA
(3)
Substituting Eq. (1) into Eq. (2) gives
vC 5 24vA and
Motion of A. You can use the constant-acceleration equations for
block A:, as
v2A 2 v2A0 5 2aA [xA 2 (xA ) 0 ]
or aA 5
v2A 2 (vA ) 20
2[xA 2 (xA ) 0 ]
(4)
a. Acceleration of A and C. You know vC and aC are down, so from
Eq. (3), you also know vA and aA are up. Substituting the given values
into Eq. (4), you find
aA 5
(180 mm/s)2 2 0
5 236 mm/s 2
2(2450 mm)
aA 5 36 mm/s2x
b
aC 5 144 mm/s2w
b
Substituting this value into aC 5 24aA, you obtain
b. Velocity and change in position of B after 2.5 s. Substituting
aA in aB 5 2aA gives
aB 5 2(236 mm/s 2 ) 5 272 mm/s 2
You can use the equations of constant acceleration to find
DvB 5 aB t 5 (272 mm/s2 )(2.5 s) 5 2180 mm/s
DxB 5 12 aB t 5 12 (272 mm/s2 )(2.5 s) 2 5 2225 mm
DvB 5 180 mm/sx b
DxB 5 225 mmx
b
REFLECT and THINK: One of the keys to solving this problem is
recognizing that since there are two cables, you need to write two
constraint equations. The directions of the answers also make sense. If
block C is accelerating downward, you would expect A and B to accelerate
upward.
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SOLVING PROBLEMS
ON YOUR OWN
I
n this section, we derived the equations that describe uniform rectilinear motion
(constant velocity) and uniformly accelerated rectilinear motion (constant
acceleration). We also introduced the concept of relative motion. We can apply the
equations for relative motion [Eqs. (11.9) through (11.11)] to the independent or
dependent motions of any two particles moving along the same straight line.
A. Independent motion of one or more particles. Organize the solution of problems
of this type as follows.
1. Begin your solution by listing the given information, sketching the system, and
selecting the origin and the positive direction of the coordinate axis [Sample Prob. 11.5].
It is always advantageous to have a visual representation of problems of this type.
2. Write the equations that describe the motions of the various particles as well as
those that describe how these motions are related [Eq. (5) of Sample Prob. 11.5].
3. Define the initial conditions, i.e., specify the state of the system corresponding
to t 5 0. This is especially important if the motions of the particles begin at different
times. In such cases, either of two approaches can be used.
a. Let t 5 0 be the time when the last particle begins to move. You must then
determine the initial position x0 and the initial velocity v0 of each of the other
particles.
b. Let t 5 0 be the time when the first particle begins to move. You must then,
in each of the equations describing the motion of another particle, replace t with
t 2 t0, where t0 is the time at which that specific particle begins to move. It is
important to recognize that the equations obtained in this way are valid only for t $ t0.
644
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B. Dependent motion of two or more particles. In problems of this type, the
particles of the system are connected to each other, typically by ropes or cables. The
method of solution of these problems is similar to that of the preceding group of
problems, except that it is now necessary to describe the physical connections between
the particles. In the following problems, the connection is provided by one or more
cables. For each cable, you will have to write equations similar to the last three
equations of Sec. 11.2C. We suggest that you use the following procedure.
1. Draw a sketch of the system and select a coordinate system, indicating clearly a
positive sense for each of the coordinate axes. For example, in Sample Probs. 11.7
and 11.8, we measured lengths downward from the upper horizontal support. It thus
follows that those displacements, velocities, and accelerations that have positive values
are directed downward.
2. Write the equation describing the constraint imposed by each cable on the
motion of the particles involved. Differentiating this equation twice, you obtain the
corresponding relations among velocities and accelerations.
3. If several directions of motion are involved, you must select a coordinate axis
and a positive sense for each of these directions. You should also try to locate the
origins of your coordinate axes so that the equations of constraints are as simple as
possible. For example, in Sample Prob. 11.7, it is easier to define the various coordinates by measuring them downward from the upper support than by measuring them
upward from the bottom support.
Finally, keep in mind that the method of analysis described in this section and the
corresponding equations can be used only for particles moving with uniform or
uniformly accelerated rectilinear motion.
645
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Problems
11.33 An airplane begins its take-off run at A with zero velocity and a
constant acceleration a. Knowing that it becomes airborne 30 s later
at B and that the distance AB is 900 m, determine (a) the acceleration a
(b) the take-off velocity vB.
A
B
Fig. P11.33
11.34 A motorist is traveling at 54 km/h when she observes that a traffic
light 240 m ahead of her turns red. The traffic light is timed to stay
red for 24 s. If the motorist wishes to pass the light without stopping
just as it turns green again, determine (a) the required uniform deceleration of the car, (b) the speed of the car as it passes the light.
54 km/h
240 m
Fig. P11.34
11.35 Steep safety ramps are built beside mountain highways to enable
vehicles with defective brakes to stop safely. A truck enters a 225-m
ramp at a high speed v0 and travels 160 m in 6 s at constant deceleration before its speed is reduced to v0 /2. Assuming the same constant deceleration, determine (a) the additional time required for the
truck to stop (b) the additional distance traveled by the truck.
v0
RS
OVE
YM
NTR
OU
C
OSS
CR
v1
Fig. P11.35
27 m
Fig. P11.36
11.36 A group of students launches a model rocket in the vertical
direction. Based on tracking data, they determine that the altitude
of the rocket was 27 m at the end of the powered portion of the
flight and that the rocket landed 16 s later. Knowing that the
descent parachute failed to deploy so that the rocket fell freely to
the ground after reaching its maximum altitude and assuming that
g 5 9.81 m/s2, determine (a) the speed v1 of the rocket at the end
of powered flight, (b) the maximum altitude reached by the rocket.
646
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11.37 A small package is released from rest at A and moves along the skate
wheel conveyor ABCD. The package has a uniform acceleration of
4.8 m/s2 as it moves down sections AB and CD, and its velocity is
constant between B and C. If the velocity of the package at D is
7.2 m/s, determine (a) the distance d between C and D, (b) the time
required for the package to reach D.
A
3m
B
3m
C
d
v
D
Fig. P11.37
11.38 A sprinter in a 100-m race accelerates uniformly for the first 35 m
and then runs with constant velocity. If the sprinter’s time for the
first 35 m is 5.4 s, determine (a) his acceleration, (b) his final
velocity, (c) his time for the race.
Fig. P11.38
11.39 Automobile A starts from O and accelerates at the constant rate of
0.75 m/s2. A short time later it is passed by bus B which is traveling
in the opposite direction at a constant speed of 6 m/s. Knowing that
bus B passes point O 20 s after automobile A started from there,
determine when and where the vehicles passed each other.
B
A
O
B
A
x
Fig. P11.39
11.40 In a boat race, boat A is leading boat B by 50 m and both boats are
traveling at a constant speed of 180 km/h. At t 5 0, the boats
accelerate at constant rates. Knowing that when B passes A, t 5 8 s
and vA 5 225 km/h, determine (a) the acceleration of A, (b) the
acceleration of B.
50 m
vB
vA
A
B
Fig. P11.40
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11.41 As relay runner A enters the 20-m-long exchange zone with a speed
of 12.9 m/s, he begins to slow down. He hands the baton to runner
B 1.82 s later as they leave the exchange zone with the same velocity.
Determine (a) the uniform acceleration of each of the runners,
(b) when runner B should begin to run.
(vA)0 = 12.9 m/s
(vB)0 = 0
A
B
11.42 Automobiles A and B are traveling in adjacent highway lanes and
at t 5 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.54 m/s2 and that B has
a constant deceleration of 0.36 m/s2, determine (a) when and
where A will overtake B, (b) the speed of each automobile at that
time.
20 m
Fig. P11.41
(vA)0 = 36 km/h
11.43 Two automobiles A and B are approaching each other in adjacent
highway lanes. At t 5 0, A and B are 1 km apart, their speeds are
vA 5 108 km/h and vB 5 63 km/h, and they are at points P and Q,
respectively. Knowing that A passes point Q 40 s after B was there
and that B passes point P 42 s after A was there, determine (a) the
uniform accelerations of A and B, (b) when the vehicles pass each
other, (c) the speed of B at that time.
(vB)0 = 54 km/h
A
B
22.5 m
x
Fig. P11.42
vA = 108 km/h
vB = 63 km/h
A
B
P
1 km
Q
Fig. P11.43
11.44 An elevator is moving upward at a constant speed of 4 m/s. A man
standing 10 m above the top of the elevator throws a ball upward
with a speed of 3 m/s. Determine (a) when the ball will hit the
elevator, (b) where the ball will hit the elevator with respect to the
location of the man.
10 m
11.45 Two rockets are launched at a fireworks display. Rocket A is launched
with an initial velocity v0 5 100 m/s and rocket B is launched
t1 seconds later with the same initial velocity. The two rockets are
timed to explode simultaneously at a height of 300 m as A is falling
and B is rising. Assuming a constant acceleration g 5 9.81 m/s2,
determine (a) the time t1, (b) the velocity of B relative to A at the time
of the explosion.
Fig. P11.44
11.46 Car A is parked along the northbound lane of a highway, and car B is
traveling in the southbound lane at a constant speed of 90 km/h. At
t 5 0, A starts and accelerates at a constant rate aA, while at t 5 5 s,
B begins to slow down with a constant deceleration of magnitude
aA /6. Knowing that when the cars pass each other x 5 90 m and
vA 5 vB, determine (a) the acceleration aA, (b) when the vehicles pass
each other, (c) the distance d between the vehicles at t 5 0.
A
v0
300 m
B
A
(vB)0 = 90 km/h
(vA)0 = 0
B
v0
x
d
Fig. P11.45
Fig. P11.46
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11.47 The elevator E shown in the figure moves downward with a constant
velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the
velocity of the counterweight W, (c) the relative velocity of the cable
C with respect to the elevator, (d) the relative velocity of the
counterweight W with respect to the elevator.
11.48 The elevator E shown starts from rest and moves upward with a
constant acceleration. If the counterweight W moves through 10 m
in 5 s, determine (a) the acceleration of the elevator and the cable C,
(b) the velocity of the elevator after 5 s.
11.49 An athlete pulls handle A to the left with a constant velocity of
0.5 m/s. Determine (a) the velocity of the weight B, (b) the relative
velocity of weight B with respect to the handle A.
W
C
E
M
Fig. P11.47 and P11.48
A
B
Fig. P11.49
11.50 An athlete pulls handle A to the left with a constant acceleration.
Knowing that after the weight B has been lifted 100 mm its velocity
is 0.6 m/s, determine (a) the accelerations of handle A and weight
B, (b) the velocity and change in position of handle A after 0.5 sec.
A
B
Fig. P11.50
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11.51 Slider block B moves to the right with a constant velocity of
300 mm/s. Determine (a) the velocity of slider block A, (b) the
velocity of portion C of the cable, (c) the velocity of portion D
of the cable, (d) the relative velocity of portion C of the cable with
respect to slider block A.
C
A
B
D
Fig. P11.51 and P11.52
11.52 At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slider block
A has moved 240 mm to the right its velocity is 60 mm/s, determine
(a) the accelerations of A and B, (b) the acceleration of portion D
of the cable, (c) the velocity and the change in position of slider
block B after 4 s.
11.53 Slider block A moves to the left with a constant velocity of 6 m/s.
Determine (a) the velocity of block B, (b) the velocity of portion D
of the cable, (c) the relative velocity of portion C of the cable with
respect to portion D.
A
C
D
B
B
100 mm/s
L
M
Fig. P11.54
Fig. P11.53
11.54 The motor M reels in the cable at a constant rate of 100 mm/s.
Determine (a) the velocity of load L, (b) the velocity of pulley B
with respect to load L.
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11.55 Collar A starts from rest and moves upward with a constant acceleration. Knowing that after 8 s the relative velocity of collar B with respect
to collar A is 0.6 m/s, determine (a) the accelerations of A and B,
(b) the velocity and the change in position of B after 6 s.
11.56 Block A starts from rest at t 5 0 and moves downward with a
constant acceleration of 150 mm/s2. Knowing that block B moves up
with a constant velocity of 75 mm/s, determine (a) the time when
the velocity of block C is zero, (b) the corresponding position of
block C.
C
A
B
A
C
Fig. P11.55
B
Fig. P11.56
C
11.57 Block B starts from rest, block A moves with a constant acceleration,
and slider block C moves to the right with a constant acceleration
of 75 mm/s2. Knowing that at t 5 2 s the velocities of B and C are
480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of
A and C, (c) the change in position of slider block C after 3 s.
11.58 Block B moves downward with a constant velocity of 20 mm/s. At
t 5 0, block A is moving upward with a constant acceleration, and
its velocity is 30 mm/s. Knowing that at t 5 3 s slider block C has
moved 57 mm to the right, determine (a) the velocity of slider block
C at t 5 0, (b) the accelerations of A and C, (c) the change in position of block A after 5 s.
11.59 The system shown starts from rest, and each component moves with
a constant acceleration. If the relative acceleration of block C with
respect to collar B is 60 mm/s2 upward and the relative acceleration
of block D with respect to block A is 110 mm/s2 downward, determine (a) the velocity of block C after 3 s, (b) the change in position
of block D after 5 s.
*11.60 The system shown starts from rest, and the length of the upper cord
is adjusted so that A, B, and C are initially at the same level. Each
component moves with a constant acceleration, and after 2 s the
relative change in position of block C with respect to block A is
280 mm upward. Knowing that when the relative velocity
of collar B with respect to block A is 80 mm/s downward, the
displacements of A and B are 160 mm downward and 320 mm
downward, respectively, determine (a) the accelerations of A and B
if aB . 10 mm/s2, (b) the change in position of block D when the
velocity of block C is 600 mm/s upward.
B
A
Fig. P11.57 and P11.58
A
B
C
D
Fig. P11.59 and P11.60
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652
Kinematics of Particles
*11.3
GRAPHICAL SOLUTIONS
In analyzing problems in rectilinear motion, it is often useful to draw
graphs of position, velocity, or acceleration versus time. Sometimes these
graphs can provide insight into the situation by indicating when quantities
increase, decrease, or stay the same. In other cases, the graphs can provide
numerical solutions when analytical methods are not available. In many
experimental situations, data are collected as a function of time, and the
methods of this section are very useful for the analysis.
x
v
a
e
op
Sl
a
x
t1
e
op
Sl
dx = v
dt
dv = a
dt
v
t
a
t1
t
t1
t
Fig. 11.10
The slope of an x–t curve at time t1 equals the velocity v at that time; the slope
of the v–t curve at time t1 equals the acceleration a at that time.
Area
t2
t1
We observed in Sec. 11.1 that the fundamental formulas
t
v
v5
v2
v2 − v1 =
v1
t2
a dt
t1
Area
t1
t2
t
x
dx
dt
and
a5
dv
dt
have a geometrical significance. The first formula says that the velocity
at any instant is equal to the slope of the x–t curve at that instant
(Fig. 11.10). The second formula states that the acceleration is equal to
the slope of the v–t curve. We can use these two properties to determine
graphically the v–t and a–t curves of a motion when the x–t curve is
known.
Integrating the two fundamental formulas from a time t1 to a time
t2, we have
t2
x2
x2 − x1 =
x1
t1
t2
t2
v dt
# v dt
t1
t2
and
v2 2 v 1 5
# a dt
Ê
(11.12)
t1
t1
t
Fig. 11.11 The area under an a–t curve
equals the change in velocity during that
time interval; the area under the v–t curve
equals the change in position during that
time interval.
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x2 2 x1 5
The first formula says that the area measured under the v−t curve from t1
to t2 is equal to the change in x during that time interval (Fig. 11.11).
Similarly, the second formula states that the area measured under the a–t
curve from t1 to t2 is equal to the change in v during that time interval.
We can use these two properties to determine graphically the x–t curve of
a motion when its v−t curve or its a–t curve is known (see Sample
Prob. 11.9).
12/10/15 11:37 AM
*11.3
Graphical Solutions
653
Graphical solutions are particularly useful when the motion considered is defined from experimental data and when x, v, and a are not
analytical functions of t. They also can be used to advantage when the
motion consists of distinct parts and when its analysis requires writing a
different equation for each of its parts. When using a graphical solution,
however, be careful to note that (1) the area under the v–t curve measures
the change in x—not x itself—and similarly, that the area under the a–t
curve measures the change in v; (2) an area above the t axis corresponds
to an increase in x or v, whereas an area located below the t axis measures
a decrease in x or v.
In drawing motion curves, it is useful to remember that, if the
velocity is constant, it is represented by a horizontal straight line; the
position coordinate x is then a linear function of t and is represented by
an oblique straight line. If the acceleration is constant and different from
zero, it is represented by a horizontal straight line; v is then a linear
function of t and is represented by an oblique straight line, and x is a
second-degree polynomial in t and is represented by a parabola. If the
acceleration is a linear function of t, the velocity and the position coordinate
are equal, respectively, to second-degree and third-degree polynomials;
a is then represented by an oblique straight line, v by a parabola, and x
by a cubic. In general, if the acceleration is a polynomial of degree n in t,
the velocity is a polynomial of degree n 1 1, and the position coordinate
is a polynomial of degree n 1 2. These polynomials are represented by
motion curves of a corresponding degree.
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654
Kinematics of Particles
Sample Problem 11.9
A particle moves in a straight line with the acceleration shown in the
figure. Knowing that it starts from the origin with v0 5 26 m/s, (a) plot
the v – t and x – t curves for 0 , t , 20 s, (b) determine its velocity, its
position, and the total distance traveled when t 5 12 s.
a(m/s2)
2
1
0
10
4
t(s)
–2
STRATEGY: You are given the graph of a versus t. You can calculate
areas under the curve to determine the v–t curve and calculate areas under
the v–t curve to determine the x–t curve.
MODELING and ANALYSIS: The particle is moving under rectilinear acceleration.
a Acceleration-Time Curve.
Initial conditions:
a(m/s 2)
t 5 0, v0 5 26 m/s, x0 5 0
Change in v 5 area under a−t curve:
2
1
v 0 = −6 m / s
10 12
0
20
t(s)
4
0 < t < 4s :
4s < t < 10 s :
10s < t < 12 s :
22
Fig. 1 Acceleration of the particle as a
function of time
12s < t < 20 s :
110
4 s < t < 5s :
16
4
0
26
22
15
5
20
t(s)
10 12
5s < t < 10 s :
10 s < t < 12 s :
210
12 s < t < 15s
Fig. 2 Velocity of the particle as a
function of time
15s < t < 20 s
v 4 = −2 m / s
2
v10 − v 4 = (2 m /s ) (6s) = + 12 m /s
v 10 = +10 m / s
2
v12 = + 6 m / s
2
v 20 = −10 m / s
v12 − v10 = (− 2 m / s )(2s) = − 4 m /s
v 20 − v12 = (−2 m / s )(8s) = −16 m /s
Change in x 5 area under v − t curve:
0 < t < 4s :
v(m/s)
2
v 4 − v 0 = (1 m /s ) (4 s ) = + 4 m /s
1
x 4 − x 0 = (−6 − 2)(4) = −16 m
2
1
x 5 − x 4 = (−2)(1) = −1 m
2
1
x 10 − x 5 = (+10)(5) = + 2 5 m
2
1
x 12 − x 10 = (+10 + 6)(2) = +16 m
2
1
x 15 − x 12 = (+ 6 )(3) = + 9 m
2
1
x 20 − x 15 = (−10)(5) = − 25 m
2
b
x0 = 0
x 4 = −16 m
x 5 = −17 m
x 10 = 8m
x 12 = + 24m
x 16 = + 33m
x 20 = + 8 m
b From above curves, you read
x(ft)
133
124
18
0
18
4 5
10 12 15
20
216 217
Fig. 3 Position of the particle as a
function of time
bee87342_ch11_615-717.indd 654
t(s)
For t512 s:
v12 5 + 6 m/s, x12 = + 24 m
Distance traveled t 5 0 to t 5 12 s
From t 5 0 s to t 5 5 s: Distance traveled 5 17 m
From t 5 5 s to t 5 12 s: Distance traveled 5 (17 + 24) = 41 m
Total distance traveled 5 58 m b
REFLECT and THINK: This problem also could have been solved
using the uniform motion equations for each interval of time that has a
different acceleration, but it would have been much more difficult and
time consuming. For a real particle, the acceleration does not instantaneously change from one value to another.
12/10/15 11:37 AM
SOLVING PROBLEMS
ON YOUR OWN
I
n this section, we reviewed and developed several graphical techniques for the
solution of problems involving rectilinear motion. These techniques can be used to
solve problems directly or to complement analytical methods of solution by providing
a visual description, and thus a better understanding, of the motion of a given body.
We suggest that you sketch one or more motion curves for several of the problems
in this section, even if these problems are not part of your homework assignment.
1. Drawing x−t, v−t, and a−t curves and applying graphical methods. We described
the following properties in Sec. 11.3, and they should be kept in mind as you use a
graphical method of solution.
a. The slopes of the x−t and v−t curves at a time t1 are equal to the velocity
and the acceleration at time t1, respectively.
b. The areas under the a−t and v−t curves between the times t1 and t2 are
equal to the change Dv in the velocity and to the change Dx in the position coordinate,
respectively, during that time interval.
c. If you know one of the motion curves, the fundamental properties we have
summarized in paragraphs a and b will enable you to construct the other two curves.
However, when using the properties of paragraph b, you must know the velocity and
the position coordinate at time t1 in order to determine the velocity and the position
coordinate at time t2. Thus, in Sample Prob. 11.9, knowing that the initial value of
the velocity was zero allowed us to find the velocity at t 5 6 s: v6 5 v0 1 Dv 5
0 1 24 ft/s 5 24 ft/s.
If you have studied the shear and bending-moment diagrams for a beam previously,
you should recognize the analogy between the three motion curves and the three
diagrams representing, respectively, the distributed load, the shear, and the bending
moment in the beam. Thus, any techniques that you have learned regarding the
construction of these diagrams can be applied when drawing the motion curves.
2. Using approximate methods. When the a–t and v–t curves are not represented
by analytical functions or when they are based on experimental data, it is often
necessary to use approximate methods to calculate the areas under these curves. In
those cases, the given area is approximated by a series of rectangles of width Dt. The
smaller the value of Dt, the smaller is the error introduced by the approximation. You
can obtain the velocity and the position coordinate from
v 5 v0 1 oaave Dt
x 5 x0 1 ovave Dt
where aave and vave are the heights of an acceleration rectangle and a velocity rectangle,
respectively.
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Problems
11.61 A particle moves in a straight line with a constant acceleration of
24 m/s2 for 6 s, zero acceleration for the next 4 s, and a constant
acceleration of 14 m/s2 for the next 4 s. Knowing that the particle
starts from the origin and that its velocity is 28 m/s during the zero
acceleration time interval, (a) construct the v–t and x–t curves for
0 # t # 14 s, (b) determine the position and the velocity of the
particle and the total distance traveled when t 5 14 s.
a (m/s2)
4
0
6
10
–4
14
t(s)
Fig. P11.61 and P11.62
11.62 A particle moves in a straight line with a constant acceleration of
24 m/s2 for 6 s, zero acceleration for the next 4 s, and a constant
acceleration of 14 m/s2 for the next 4 s. Knowing that the particle
starts from the origin with v0 5 16 m/s, (a) construct the v–t and
x–t curves for 0 # t # 14 s, (b) determine the amount of time during
which the particle is further than 16 m from the origin.
11.63 A particle moves in a straight line with the velocity shown in the
figure. Knowing that x 5 2540 m at t 5 0, (a) construct the a–t
and x–t curves for 0 , t , 50 s, and determine (b) the total distance
traveled by the particle when t 5 50 s, (c) the two times at which
x 5 0.
v (m/s)
60
–5
–20
26
10
41
46
t (s)
Fig. P11.63 and P11.64
11.64 A particle moves in a straight line with the velocity shown in the
figure. Knowing that x 5 2540 m at t 5 0, (a) construct the a–t
and x–t curves for 0 , t , 50 s, and determine (b) the maximum
value of the position coordinate of the particle, (c) the values of t for
which the particle is at x 5 100 m.
v (m/s)
18
6
0
24 30
10
–18
Fig. P11.65
18
t (s)
11.65 A particle moves in a straight line with the velocity shown in the
figure. Knowing that x 5 248 m at t 5 0, draw the a–t and x–t
curves for 0 , t , 40 s and determine (a) the maximum value of
the position coordinate of the particle, (b) the values of t for which
the particle is at a distance of 108 m from the origin.
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11.66 A parachutist is in free fall at a rate of 200 km/h when he opens his
parachute at an altitude of 600 m. Following a rapid and constant
deceleration, he then descends at a constant rate of 50 km/h from
586 m to 30 m, where he maneuvers the parachute into the wind to
further slow his descent. Knowing that the parachutist lands with a
negligible downward velocity, determine (a) the time required for
the parachutist to land after opening his parachute, (b) the initial
deceleration.
v
11.67 A commuter train traveling at 60 km/h is 4.5 km from a station. The
train then decelerates so that its speed is 30 km/h when it is 0.75 km
from the station. Knowing that the train arrives at the station 7.5
min after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train to travel the first
3.75 km, (b) the speed of the train as it arrives at the station, (c) the
final constant deceleration of the train.
60 km/h
4.5 km
Fig. P11.66
Fig. P11.67
11.68 A temperature sensor is attached to slider AB which moves back
and forth through 1500 mm. The maximum velocities of the slider
are 300 mm/s to the right and 750 mm/s to the left. When the slider
is moving to the right, it accelerates and decelerates at a constant
rate of 150 mm/s2; when moving to the left, the slider accelerates
and decelerates at a constant rate of 500 mm/s2. Determine the time
required for the slider to complete a full cycle, and construct the
v–t and x–t curves of its motion.
11.69 In a water-tank test involving the launching of a small model boat,
the model’s initial horizontal velocity is 6 m/s and its horizontal
acceleration varies linearly from 212 m/s2 at t 5 0 to 22 m/s2 at
t 5 t1 and then remains equal to 22 m/s2 until t 5 1.4 s. Knowing
that v 5 1.8 m/s when t 5 t1, determine (a) the value of t1, (b) the
velocity and the position of the model at t 5 1.4 s.
11.70 The acceleration record shown was obtained for a small airplane
traveling along a straight course. Knowing that x 5 0 and v 5
60 m/s when t 5 0, determine (a) the velocity and position of
the plane at t 5 20 s, (b) its average velocity during the interval
6 s , t , 14 s.
11.71 In a 400-m race, runner A reaches her maximum velocity vA in
4 s with constant acceleration and maintains that velocity until she
reaches the halfway point with a split time of 25 s. Runner B reaches
her maximum velocity vB in 5 s with constant acceleration and maintains that velocity until she reaches the halfway point with a split
time of 25.2 s. Both runners then run the second half of the race
with the same constant deceleration of 0.1 m/s2. Determine (a) the
race times for both runners, (b) the position of the winner relative
to the loser when the winner reaches the finish line.
x
1500 mm
A
B
Fig. P11.68
x
v0 = 6 m/s
Fig. P11.69
a (m/s2)
0.75
6
0
8 10
12 14
20 t(s)
–0.75
Fig. P11.70
B
A
200 m
200 m
Fig. P11.71
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11.72 A car and a truck are both traveling at the constant speed of 50 km/h;
the car is 12 m behind the truck. The driver of the car wants to
pass the truck, i.e., he wishes to place his car at B, 12 m in front
of the truck, and then resume the speed of 50 km/h. The maximum
acceleration of the car is 1.5 m/s2 and the maximum deceleration
obtained by applying the brakes is 6 m/s2. What is the shortest time
in which the driver of the car can complete the passing operation
if he does not at any time exceed a speed of 75 km/h? Draw the
v–t curve.
A
4.8 m
B
12 m
15 m
12 m
Fig. P11.72
11.73 Solve Prob. 11.72, assuming that the driver of the car does not pay
any attention to the speed limit while passing and concentrates on
reaching position B and resuming a speed of 50 km/h in the shortest
possible time. What is the maximum speed reached? Draw the
v–t curve.
11.74 Car A is traveling on a highway at a constant speed (vA)0 5 90 km/h
and is 120 m from the entrance of an access ramp when car B enters
the acceleration lane at that point at a speed (vB)0 5 25 km/h.
Car B accelerates uniformly and enters the main traffic lane after traveling 60 m in 5 s. It then continues to accelerate at the same rate until
it reaches a speed of 90 km/h, which it then maintains. Determine the
final distance between the two cars.
120 m
A
(vA)0
B
12 m
Fig. P11.75
(vB)0
Fig. P11.74
11.75 An elevator starts from rest and moves upward, accelerating at a rate
of 1.2 m/s2 until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing
12 m above the initial position of the top of the elevator throws a
ball upward with an initial velocity of 20 m/s. Determine when the
ball will hit the elevator.
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11.76 Car A is traveling at 60 km/h when it enters a 40 km/h speed zone. The
driver of car A decelerates at a rate of 5 m/s2 until reaching a speed
of 40 km/h, which she then maintains. When car B, which was initially
20 m behind car A and traveling at a constant speed of 70 km/h, enters
the speed zone, its driver decelerates at a rate of 6 m/s2 until reaching
a speed of 35 km/h. Knowing that the driver of car B maintains a speed
of 35 km/h, determine (a) the closest that car B comes to car A, (b) the
time at which car A is 25 m in front of car B.
(vB)0 = 70 km/h
(vA)0 = 60 km/h
B
A
20 m
Fig. P11.76
11.77 An accelerometer record for the motion of a given part of a mechanism is approximated by an arc of a parabola for 0.2 s and a straight
line for the next 0.2 s as shown in the figure. Knowing that v 5 0
when t 5 0 and x 5 0.4 m when t 5 0.4 s, (a) construct the v–t
curve for 0 # t # 0.4 s, (b) determine the position of the part at
t 5 0.3 s and t 5 0.2 s.
a (m/s2)
a = 12 – 100t2
12
8
0
a = 16 – 40t
0
0.2
0.4 t (s)
Fig. P11.77
11.78 A car is traveling at a constant speed of 54 km/h when its driver
sees a child run into the road. The driver applies her brakes until the
child returns to the sidewalk and then accelerates to resume her
original speed of 54 km/h; the acceleration record of the car is shown
in the figure. Assuming x 5 0 when t 5 0, determine (a) the time
t1 at which the velocity is again 54 km/h, (b) the position of the car
at that time, (c) the average velocity of the car during the interval
1 s # t # t1.
a (m/s2)
2
0
1
2
4.5
t1 t(s)
–6
Fig. P11.78
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11.79 An airport shuttle train travels between two terminals that are
2.5 km apart. To maintain passenger comfort, the acceleration of the
train is limited to 61.2 m/s2, and the jerk, or rate of change of
acceleration, is limited to 60.24 m/s2 per second. If the shuttle has
a maximum speed of 30 km/h, determine (a) the shortest time for
the shuttle to travel between the two terminals, (b) the corresponding
average velocity of the shuttle.
11.80 During a manufacturing process, a conveyor belt starts from rest and
travels a total of 400 mm before temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to
61.5 m/s2 per second, determine (a) the shortest time required for
the belt to move 400 mm, (b) the maximum and average values of
the velocity of the belt during that time.
11.81 Two seconds are required to bring the piston rod of an air cylinder to
rest; the acceleration record of the piston rod during the 2 s is as
shown. Determine by approximate means (a) the initial velocity of
the piston rod, (b) the distance traveled by the piston rod as it is
brought to rest.
–a (m/s2)
4.0
3.0
2.0
1.0
0
0
0.25
0.5
0.75
1.0
1.25
1.5
1.75
2.0
t (s)
Fig. P11.81
11.82 The acceleration record shown was obtained during the speed trials
of a sports car. Knowing that the car starts from rest, determine by
approximate means (a) the velocity of the car at t 5 8 s, (b) the
distance the car has traveled at t 5 20 s.
a (m/s2)
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0
0
2
4
6
8
10
12
14
16
18
20
22
t (s)
Fig. P11.82
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11.83 A training airplane has a velocity of 38 m/s when it lands on an
aircraft carrier. As the arresting gear of the carrier brings the airplane
to rest, the velocity and the acceleration of the airplane are recorded;
the results are shown (solid curve) in the figure. Determine by
approximate means (a) the time required for the airplane to come to
rest, (b) the distance traveled in that time.
–a (m/s2)
18
15
12
9
6
3
0
0
6
12
18
24
30
36
v (m/s)
42
Fig. P11.83
11.84 Shown in the figure is a portion of the experimentally determined v–x
curve for a shuttle cart. Determine by approximate means the acceleration of the cart when (a) x 5 250 mm, (b) v 5 2000 mm/s.
v (mm/s)
2500
2000
1500
1000
500
0
0
250
500
750
1000
1250
x (mm)
Fig. P11.84
11.85 An elevator starts from rest and rises 40 m to its maximum velocity
in T s with the acceleration record shown in the figure. Determine
(a) the required time T, (b) the maximum velocity, (c) the velocity
and position of the elevator at t 5 T/2.
a (m/s2)
0.6
0
T/3
T
t(s)
Fig. P11.85
661
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11.86 The acceleration of an object subjected to the pressure wave of a large
explosion is defined approximately by the curve shown. The object is
initially at rest and is again at rest at time t1. Using the method of Sec.
11.8, determine (a) the time t1, (b) the distance through which the
object is moved by the pressure wave.
a (m/s2)
30
t(s)
–10
0.8 s
t1
Fig. P11.86
11.87 As shown in the figure, from t 5 0 to t 5 4 s, the acceleration of a
given particle is represented by a parabola. Knowing that x 5 0 and
v 5 8 m/s when t 5 0, (a) construct the v–t and x–t curves for 0 , t
, 4 s, (b) determine the position of the particle at t 5 3 s. (Hint:
Use table inside the front cover.)
a (m/s2)
2
4
t(s)
a = – 3 (t – 2)2 m/s2
–12
Fig. P11.87
11.88 A particle moves in a straight line with the acceleration shown in
the figure. Knowing that the particle starts from the origin with
v0 5 22 m/s, (a) construct the v–t and x–t curves for 0 , t , 18 s,
(b) determine the position and the velocity of the particle and the
total distance traveled when t 5 18 s.
a (m/s2)
6
2
8
–0.75
12
t(s)
Fig. P11.88
662
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11.4
11.4
CURVILINEAR MOTION
OF PARTICLES
When a particle moves along a curve other than a straight line, we say
that the particle is in curvilinear motion. We can use position, velocity,
and acceleration to describe the motion, but now we must treat these
quantities as vectors because they can have directions in two or three
dimensions.
11.4A
663
Curvilinear Motion of Particles
Dr
Dt
(11.13)
As Dt and Dr become shorter, the points P and P9 get closer together.
Thus, the vector v obtained in the limit must be tangent to the path of the
particle (Fig. 11.12c).
Because the position vector r depends upon the time t, we can refer
to it as a vector function of the scalar variable t and denote it by r(t).
Extending the concept of the derivative of a scalar function introduced in
elementary calculus, we refer to the limit of the quotient Dr/Dt as the
derivative of the vector function r(t). We have
Velocity vector
v5
dr
dt
∆s
∆r
r'
P
r
x
(a)
To define the position P occupied by a particle in curvilinear motion at a
given time t, we select a fixed reference system, such as the x, y, z axes
shown in Fig. 11.12a, and draw the vector r joining the origin O and
point P. The vector r is characterized by its magnitude r and its direction
with respect to the reference axes, so it completely defines the position of
the particle with respect to those axes. We refer to vector r as the position
vector of the particle at time t.
Consider now the vector r9 defining the position P9 occupied by the
same particle at a later time t 1 Dt. The vector Dr joining P and P9
represents the change in the position vector during the time interval Dt
and is called the displacement vector. We can check this directly from
Fig. 11.12a, where we obtain the vector r9 by adding the vectors r and
Dr according to the triangle rule. Note that Dr represents a change in
direction as well as a change in magnitude of the position vector r.
We define the average velocity of the particle over the time interval
Dt as the quotient of Dr and Dt. Since Dr is a vector and Dt is a scalar,
the quotient Dr/Dt is a vector attached at P with the same direction as Dr
and a magnitude equal to the magnitude of Dr divided by Dt (Fig. 11.12b).
We obtain the instantaneous velocity of the particle at time t by
taking the limit as the time interval Dt approaches zero. The instantaneous
velocity is thus represented by the vector
Dt y0
P'
O
Position, Velocity, and
Acceleration Vectors
v 5 lim
y
z
∆r
∆t
y
P'
r'
P
O
r
x
(b)
z
y
v
P
r
O
s
P0
x
(c)
z
Fig. 11.12 (a) Position vectors for a particle
moving along a curve from P to P9; (b) the
average velocity vector is the quotient of the
change in position to the elapsed time
interval; (c) the instantaneous velocity vector
is tangent to the particle’s path.
(11.14)
The magnitude v of the vector v is called the speed of the particle.
We can obtain the speed by substituting the magnitude of this vector,
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664
Kinematics of Particles
y
y'
v'
Q'
∆v
v
Q
P'
v'
v
P
O
O'
x
(b)
z'
(a)
z
x'
y'
a
which is represented by the straight-line segment PP9, for the vector Dr
in formula (11.13). However, the length of segment PP9 approaches the
length Ds of arc PP9 as Dt decreases (Fig. 11.12a). Therefore, we can
write
Q
Hodograph
v 5 lim
v
Dt y0
O'
x'
(c)
z'
y
Path
v
a
P
r
O
Ds
PP9
5 lim
Dt y0 Dt
Dt
x
(d)
Fig. 11.13 (a) Velocities v and v9 of a
particle at two different times; (b) the vector
change in the particle’s velocity during the
time interval; (c) the instantaneous
acceleration vector is tangent to the
hodograph; (d) in general, the acceleration
vector is not tangent to the particle’s path.
bee87342_ch11_615-717.indd 664
d
ds
dt
(11.15)
Thus, we obtain the speed v by finding the length s of the arc described
by the particle and differentiating it with respect to t.
Now let’s consider the velocity v of the particle at time t and its
velocity v9 at a later time t 1 Dt (Fig. 11.13a). Let us draw both vectors
v and v9 from the same origin O9 (Fig. 11.13b). The vector Dv joining Q
and Q9 represents the change in the velocity of the particle during the time
interval Dt, since we can obtain the vector v9 by adding the vectors v and
Dv. Again, note that Dv represents a change in the direction of the velocity
as well as a change in speed. We define the average acceleration of the
particle over the time interval Dt as the quotient of Dv and Dt. Since Dv
is a vector and Dt is a scalar, the quotient Dv/Dt is a vector in the same
direction as Dv.
We obtain the instantaneous acceleration of the particle at time t
by choosing increasingly smaller values for Dt and Dv. The instantaneous
acceleration is thus represented by the vector
a 5 lim
Dt y0
z
v5
Dv
Dt
(11.16)
Noting that the velocity v is a vector function v(t) of the time t, we can
refer to the limit of the quotient Dv/Dt as the derivative of v with respect
to t. We have
Acceleration vector
a5
d
dv
dt
(11.17)
12/10/15 11:37 AM
11.4
Observe that the acceleration a is tangent to the curve described by
the tip Q of the vector v when we draw v from a fixed origin O9
(Fig. 11.13c). However, in general, the acceleration is not tangent to the
path of the particle (Fig. 11.13d). The curve described by the tip of v and
shown in Fig. 11.13c is called the hodograph of the motion.
11.4B
Curvilinear Motion of Particles
665
y
∆P
P(u + ∆u)
Derivatives of Vector Functions
We have just seen that we can represent the velocity v of a particle in
curvilinear motion by the derivative of the vector function r(t) characterizing
the position of the particle. Similarly, we can represent the acceleration a
of the particle by the derivative of the vector function v(t). Here we give
a formal definition of the derivative of a vector function and establish a
few rules governing the differentiation of sums and products of vector
functions.
Let P(u) be a vector function of the scalar variable u. By that, we
mean that the scalar u completely defines the magnitude and direction of
the vector P. If the vector P is drawn from a fixed origin O and the scalar
u is allowed to vary, the tip of P describes a given curve in space. Consider
the vectors P corresponding, respectively, to the values u and u 1 Du
of the scalar variable (Fig. 11.14a). Let DP be the vector joining the tips
of the two given vectors. Then we have
P(u)
O
x
(a)
z
y
dP
du
DP 5 P(u 1 Du) 2 P(u)
Dividing through by Du and letting Du approach zero, we define the
derivative of the vector function P(u) as
P(u 1 Du) 2 P(u)
dP
DP
5 lim
5 lim
du
Du y0 Du
Du y0
Du
P(u)
O
x
(11.18)
As Du approaches zero, the line of action of DP becomes tangent to the
curve of Fig. 11.14a. Thus, the derivative dP/du of the vector function
P(u) is tangent to the curve described by the tip of P(u) (Fig. 11.14b).
The standard rules for the differentiation of the sums and products
of scalar functions extend to vector functions. Consider first the sum of
two vector functions P(u) and Q(u) of the same scalar variable u.
According to the definition given in Eq. (11.18), the derivative of the
vector P 1 Q is
z
(b)
Fig. 11.14 (a) The change in vector function
for a particle moving along a curvilinear
path; (b) the derivative of the vector function
is tangent to the path described by the tip of
the function.
d(P 1 Q)
D (P 1 Q)
DQ
DP
5 lim
5 lim a
1
b
du
Du
Du y0
Du y0 Du
Du
or since the limit of a sum is equal to the sum of the limits of its terms,
d(P 1 Q)
DQ
DP
5 lim
1 lim
Du y0 Du
Du y0 Du
du
d(P 1 Q)
dQ
dP
d
5
1
du
du
du
(11.19)
That is, the derivative of a sum of vector functions equals the sum of the
derivative of each function separately.
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666
Kinematics of Particles
We now consider the product of a scalar function f(u) and a
vector function P(u) of the same scalar variable u. The derivative of the
vector f P is
d( f P)
( f 1 D f )(P 1 DP) 2 f P
Df
DP
5 lim
5 lim a P 1 f
b
du
Du y0
Du
Du y0 Du
Du
or recalling the properties of the limits of sums and products,
d(f
( P)
(f
d
df
d
dP
5
P1f
du
du
du
(11.20)
In a similar way, we can obtain the derivatives of the scalar product and
the vector product of two vector functions P(u) and Q(u). Thus,
d(P ? Q)
dQ
d
dP
5
?Q1P?
du
du
du
(11.21)
d(P 3 Q)
dQ
d
dP
5
3Q1P3
du
du
du
(11.22)†
We can use the properties just established to determine the rectangular components of the derivative of a vector function P(u). Resolving
P into components along fixed rectangular axes x, y, and z, we have
P 5 Pxi 1 Py j 1 Pzk
(11.23)
where Px, Py, and Pz are the rectangular scalar components of the vector
P, and i, j, and k are the unit vectors corresponding, respectively, to the
x, y, and z axes (Sec. 2.12 or Appendix A). From Eq. (11.19), the derivative of P is equal to the sum of the derivatives of the terms in the righthand side. Since each of these terms is the product of a scalar and a vector
function, we should use Eq. (11.20). However, the unit vectors i, j, and k
have a constant magnitude (equal to 1) and fixed directions. Their derivatives are therefore zero, and we obtain
dPy
dP
d z
dP
dPx
dP
d
dP
5
i1
j1
k
du
du
du
du
(11.24)
Note that the coefficients of the unit vectors are, by definition, the scalar
components of the vector dP/du. We conclude that we can obtain the
rectangular scalar components of the derivative dP/du of the vector function
P(u) by differentiating the corresponding scalar components of P.
Rate of Change of a Vector. When the vector P is a function of
the time t, its derivative dP/dt represents the rate of change of P with
respect to the frame Oxyz. Resolving P into rectangular components and
using Eq. (11.24), we have
dPy
dPz
dPx
dP
5
i1
j1
k
dt
dt
dt
dt
†
Since the vector product is not commutative (see Sec. 3.4), the order of the factors in
Eq. (11.22) must be maintained.
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11.4
Curvilinear Motion of Particles
Alternatively, using dots to indicate differentiation with respect to t gives
Ṗ 5 Ṗxi 1 Ṗy j 1 Ṗzk
y'
P(t)
(11.249)
As you will see in Sec. 15.5, the rate of change of a vector as
observed from a moving frame of reference is, in general, different from
its rate of change as observed from a fixed frame of reference. However,
if the moving frame O9x9y9z9 is in translation, i.e., if its axes remain
parallel to the corresponding axes of the fixed frame Oxyz (Fig. 11.15),
we can use the same unit vectors i, j, and k in both frames, and at any
given instant, the vector P has the same components Px, Py, and Pz in both
frames. It follows from Eq. (11.249) that the rate of change Ṗ is the same
with respect to the frames Oxyz and O9x9y9z9. Therefore,
The rate of change of a vector is the same with respect to a fixed
frame and with respect to a frame in translation.
667
y
O'
x'
O
x
z'
z
Fig. 11.15 The rate of change of a vector is
the same with respect to a fixed frame of
reference and with respect to a frame in
translation.
This property will greatly simplify our work, since we will be concerned
mainly with frames in translation.
11.4C
Rectangular Components
of Velocity and Acceleration
y
Suppose the position of a particle P is defined at any instant by its
rectangular coordinates x, y, and z. In this case, it is often convenient to
resolve the velocity v and the acceleration a of the particle into rectangular
components (Fig. 11.16).
To resolve the position vector r of the particle into rectangular
components, we write
r 5 xi 1 yj 1 zk
vy
v
P
j
(11.25)
Here the coordinates x, y, and z are functions of t. Differentiating twice,
we obtain
Velocity and acceleration in
rectangular components
yj
r
xi
O
x
i
zk
k
(a)
ay
y
(11.26)
a
(11.27)
P
where ẋ , ẏ , and ż and ẍ, ÿ, and z̈ represent, respectively, the first and second
derivatives of x, y, and z with respect to t. It follows from Eqs. (11.26)
and (11.27) that the scalar components of the velocity and acceleration are
vx 5 ẋ
vy 5 ẏ
vz 5 ż
(11.28)
ax 5 ẍ
ay 5 ÿ
az 5 z̈
(11.29)
A positive value for vx indicates that the vector component vx is directed to
the right, and a negative value indicates that it is directed to the left. The sense
of each of the other vector components is determined in a similar way from
the sign of the corresponding scalar component. If desired, we can obtain the
magnitudes and directions of the velocity and acceleration from their scalar
components using the methods of Secs. 2.2A and 2.4A (or Appendix A).
bee87342_ch11_615-717.indd 667
vz
z
dr
.
.
.
5 x i 1 yyjj 1 z k
v5
dt
d
dv
$
$
$
a5
5 xi 1 yj 1 z k
dt
vx
j
O
ax
az
r
x
i
k
z
(b)
Fig. 11.16 (a) Rectangular components of
position and velocity for a particle P;
(b) rectangular components of acceleration
for particle P.
12/10/15 11:37 AM
668
Kinematics of Particles
The use of rectangular components to describe the position, velocity,
and acceleration of a particle is particularly effective when the component
ax of the acceleration depends only upon t, x, and/or vx, and similarly when
ay depends only upon t, y, and/or vy, and when az depends upon t, z, and/or
vz. In this case, we can integrate Equations (11.28) and (11.29) independently.
In other words, the motion of the particle in the x direction, its motion in
the y direction, and its motion in the z direction can be treated separately.
In the case of the motion of a projectile, we can show (see
Sec. 12.1D) that the components of the acceleration are
ax 5 ẍ 5 0
Photo 11.3 The motion of this snowboarder
in the air is a parabola, assuming we can
neglect air resistance.
v0
vy 5 ẏ 5 (vy)0 2 gt
y 5 y0 1 (vy)0t 2 12 gt2
vz 5 ż 5 (vz)0
z 5 z0 1 (vz)0t
If the projectile is fired in the xy plane from the origin O, we have
x0 5 y0 5 z0 5 0 and (vz)0 5 0, so the equations of motion reduce to
y
(vx)0
vx 5 (vx)0
x 5 (vx)0t
O
x
(a) Motion of a projectile
y
(vy)0
az 5 z̈ 5 0
if the resistance of the air is neglected. Denoting the coordinates of a gun
by x0, y0, and z0 and the components of the initial velocity v0 of the
projectile by (vx)0, (vy)0, and (vz)0, we can integrate twice in t and obtain
vx 5 ẋ 5 (vx)0
x 5 x0 1 (vx)0t
(vy)0
ay 5 ÿ 5 2g
(vx)0
(vx)0
x
(b) Equivalent rectilinear motions
Fig. 11.17 The motion of a projectile
(a) consists of uniform horizontal motion and
uniformly accelerated vertical motion and
(b) is equivalent to two independent
rectilinear motions.
vy 5 (vy)0 2 gt
y 5 (vy)0t 2 12 gt2
vz 5 0
z50
These equations show that the projectile remains in the xy plane, that its
motion in the horizontal direction is uniform, and that its motion in the
vertical direction is uniformly accelerated. Thus, we can replace the motion
of a projectile by two independent rectilinear motions, which are easily
visualized if we assume that the projectile is fired vertically with an initial
velocity (vy)0 from a platform moving with a constant horizontal velocity
(vx)0 (Fig. 11.17). The coordinate x of the projectile is equal at any instant
to the distance traveled by the platform, and we can compute its
coordinate y as if the projectile were moving along a vertical line.
Additionally, because the (vx)0 values are the same, the projectile will land
on the platform regardless of the value of (vy)0.
Note that the equations defining the coordinates x and y of a
projectile at any instant are the parametric equations of a parabola. Thus,
the trajectory of a projectile is parabolic. This result, however, ceases to
be valid if we take into account the resistance of the air or the variation
with altitude of the acceleration due to gravity.
11.4D
Motion Relative to a Frame
in Translation
We have just seen how to describe the motion of a particle by using a single
frame of reference. In most cases, this frame was attached to the earth and
was considered to be fixed. Now we want to analyze situations in which
it is convenient to use several frames of reference simultaneously. If one
of the frames is attached to the earth, it is called a fixed frame of reference,
and the other frames are referred to as moving frames of reference. You
should recognize, however, that the selection of a fixed frame of reference
is purely arbitrary. Any frame can be designated as “fixed”; all other frames
not rigidly attached to this frame are then described as “moving.”
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11.4
Consider two particles A and B moving in space (Fig. 11.18). The
vectors rA and rB define their positions at any given instant with respect
to the fixed frame of reference Oxyz. Consider now a system of axes x9,
y9, and z9 centered at A and parallel to the x, y, and z axes. Suppose that,
while the origin of these axes moves, their orientation remains the same;
then the frame of reference Ax9y9z9 is in translation with respect to Oxyz.
The vector rB/A joining A and B defines the position of B relative to the
moving frame Ax9y9z9 (or for short, the position of B relative to A).
Figure 11.18 shows that the position vector rB of particle B is the
sum of the position vector rA of particle A and of the position vector rB/A
of B relative to A; that is,
Relative
position
rB 5 rA 1 rB/A
Curvilinear Motion of Particles
669
y'
y
B
rB
rB/A
rA
x'
A
O
x
z'
z
Fig. 11.18
The vector rB/A defines the
position of B with respect to moving
frame A.
(11.30)
Differentiating Eq. (11.30) with respect to t within the fixed frame of
reference, and using dots to indicate time derivatives, we have
ṙB 5 ṙA 1 ṙB/A
(11.31)
The derivatives ṙA and ṙB represent, respectively, the velocities vA and vB
of the particles A and B. Since Ax9y9z9 is in translation, the derivative ṙB/A
represents the rate of change of rB/A with respect to the frame Ax9y9z9 as
well as with respect to the fixed frame (Sec. 11.4B). This derivative, therefore, defines the velocity vB/A of B relative to the frame Ax9y9z9 (or for
short, the velocity vB/A of B relative to A). We have
Relative
velocity
vB 5 vA 1 vB/A
(11.32)
Vhelicopter
Vhelicopter/ship
Vship
Photo 11.4 The pilot of a helicopter landing
on a moving carrier must take into account
the relative motion of the ship.
Differentiating Eq. (11.32) with respect to t, and using the derivative v̇B/A
to define the acceleration aB/A of B relative to the frame Ax9y9z9 (or for
short, the acceleration aB/A of B relative to A), we obtain
Relative
acceleration
aB 5 aA 1 aB/A
(11.33)
We refer to the motion of B with respect to the fixed frame Oxyz as the
absolute motion of B. The equations derived in this section show that we
can obtain the absolute motion of B by combining the motion of A and
the relative motion of B with respect to the moving frame attached to A.
Equation (11.32), for example, expresses that the absolute velocity vB of
particle B can be obtained by vectorially adding the velocity of A and the
velocity of B relative to the frame Ax9y9z9. Equation (11.33) expresses a
similar property in terms of the accelerations. (Note that the product of the
subscripts A and B/A used in the right-hand sides of Eqs. (11.30) through
(11.33) is equal to the subscript B used in their left-hand sides.) Keep in mind,
however, that the frame Ax9y9z9 is in translation; that is, while it moves with A, it
maintains the same orientation. As you will see later (Sec. 15.7), you must
use different relations in the case of a rotating frame of reference.
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670
Kinematics of Particles
Sample Problem 11.10
A projectile is fired from the edge of a 150-m cliff with an initial velocity
of 180 m/s at an angle of 30° with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the
projectile strikes the ground, (b) the greatest elevation above the ground
reached by the projectile.
180 m/s
30°
150 m
STRATEGY: This is a projectile motion problem, so you can consider
the vertical and horizontal motions separately. First determine the equations
governing each direction, and then use them to find the distances.
x
y
180 m/s
(vy)0
O
MODELING and ANALYSIS: Model the projectile as a particle and
neglect the effects of air resistance. The vertical motion has a constant
acceleration. Choosing the positive sense of the y axis upward and placing
the origin O at the gun (Fig. 1), you have
a = –9.81 m /s 2
30°
(vy)0 5 (180 m/s) sin 30° 5 190 m/s
a 5 29.81 m/s2
–150 m
Fig. 1 Acceleration and initial
velocity of the projectile in the
y-direction.
30°
(vx)0
vy 5 (vy)0 1 at
y 5 (vy)0 t 1 12 at2
v2y 5 (vy)20 1 2ay
(1)
(2)
(3)
vy 5 90 2 9.81t
y 5 90t 2 4.90t2
v2y 5 8100 2 19.62y
The horizontal motion has zero acceleration. Choose the positive sense of
the x axis to the right (Fig. 2), which gives you
180 m/s
O
Substitute these values into the equations for motion with constant acceleration. Thus,
x
(vx)0 5 (180 m/s) cos 30° 5 1155.9 m/s
Substituting into the equation for constant acceleration, you obtain
Fig. 2 Initial velocity of the
projectile in the x-direction.
x 5 (vx)0 t
(4)
x 5 155.9t
a. Horizontal Distance. When the projectile strikes the ground,
y 5 2150 m
Substituting this value into Eq. (2) for the vertical motion, you have
2150 5 90t 2 4.90t2
t2 2 18.37t 2 30.6 5 0
t 5 19.91 s
Substituting t 5 19.91 s into Eq. (4) for the horizontal motion, you obtain
x 5 155.9(19.91)
x 5 3100 m
b
b. Greatest Elevation. When the projectile reaches its greatest elevation,
vy 5 0; substituting this value into Eq. (3) for the vertical motion, you have
0 5 8100 2 19.62y
y 5 413 m
Greatest elevation above ground 5 150 m 1 413 m 5 563 m b
REFLECT and THINK: Because there is no air resistance, you can treat
the vertical and horizontal motions separately and can immediately write
down the algebraic equations of motion. If you did want to include air
resistance, you must know the acceleration as a function of the speed (you
will see how to derive this in Chapter 12), and then you need to use the
basic kinematic relationships, separate variables, and integrate.
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11.4
671
Curvilinear Motion of Particles
Sample Problem 11.11
240 m/s
A
B
600 m
a
3600 m
v0 = 240 m/s
O
a
STRATEGY: This is a projectile motion problem, so you can consider the
vertical and horizontal motions separately. First determine the equations governing the motion in each direction, and then use them to find the firing angle.
B
(vx)0 = 240 cos A projectile is fired with an initial velocity of 240 m/s at a target B located
600 m above the gun A and at a horizontal distance of 3,600 m. Neglecting air resistance, determine the value of the firing angle α needed to hit
the target.
MODELING and ANALYSIS:
x
3,600 m
Horizontal Motion.
Place the origin of the coordinate axes at the
gun (Fig. 1). Then
(vx)0 5 240 cos α
Fig. 1 Initial velocity of the
Substituting into the equation of uniform horizontal motion, you obtain
projectile in the x-direction.
x 5 (240 cos α)t
x 5 (vx)0 t
Obtain the time required for the projectile to move through a horizontal
distance of 3,600 m by setting x equal to 3,600 m.
y
3600 5 (240 cos α)t
3,600
15
t5
5
240 cos α
cos α
a = – 9.81 m/s2
Vertical Motion.
B
O
a
v0 = 240 m/s
600 m
(vy)0 = 240 sin a
Fig. 2 Acceleration and initial
velocity of the projectile in the
y-direction.
Again, place the origin at the gun (Fig. 2).
(vy)0 5 240 sin α
a 5 29.81 m/s2
Substituting into the equation for constant acceleration in the vertical
direction, you obtain
y 5 (vy)0 t 1 12 at2
y 5 (240 sin α)t 2 4.905t2
Projectile Hits Target. When x 5 3600 m, you want y 5 600 m.
Substituting for y and setting t equal to the value found previously, you have
600 5 240 sin a
15
15
24.905
cos a
cos a
2
(1)
Since 1/cos2 a 5 sec2 a 5 1 1 tan2 a, we have
600 5 240(15) tan a 2 4.905(152)(1 1 tan2 a)
1104 tan2 a 2 3600 tan a 1 1704 5 0
Solving this quadratic equation for tan a, we have
tan a 5 0.575
and
tan a 5 2.69
a 5 29.9°
and
a 5 69.6°
b
The target will be hit if either of these two firing angles is used (Fig. 3).
69.6˚
A
B
29.9˚
Fig. 3 Firing angles that will
hit target B.
bee87342_ch11_615-717.indd 671
REFLECT and THINK: It is a well-known characteristic of projectile
motion that you can hit the same target by using either of two firing
angles. We used trigonometry to write the equation in terms of tan α, but
most calculators or computer programs like Maple, Matlab, or Mathematica
also can be used to solve (1) for α. You must be careful when using these
tools, however, to make sure that you find both angles.
12/10/15 11:38 AM
672
Kinematics of Particles
Sample Problem 11.12
20°
v0
1m
45°
d
y
v0
A conveyor belt at an angle of 20º with the horizontal is used to transfer small packages to other parts of an industrial plant. A worker tosses
a package with an initial velocity v0 at an angle of 45º so that its
velocity is parallel to the belt as it lands 1 m above the release point.
Determine (a) the magnitude of v0, (b) the horizontal distance d.
STRATEGY: This is a projectile motion problem, so you can consider the vertical and the horizontal motions separately. First determine the equations governing the motion in each direction, then use
them to determine the unknown quantities.
MODELING and ANALYSIS:
45°
O
x
Horizontal Motion. Placing the axes of your origin at the location
where the package leaves the workers hands (Fig. 1), you can write
Horizontal: vx 5 v0 cos 45°
Fig. 1 Initial velocity
of the package.
and
Vertical: vy 5 v0 sin 45° 2 gt
and
x 5 (v0 cos 45°) t
y 5 (v0 sin 45°) t 2
1 2
gt
2
Landing on the Belt. The problem statement indicates that
when the package lands on the belt, its velocity vector will be in the
same direction as the belt is moving. If this happens when t 5 t1,
you can write
vy
vx
5 tan 20° 5
v0 sin 45° 2 gt1
gt1
512
v0 cos 45°
v0 cos 45°
(1)
This equation has two unknown quantities: t1 and v0. Therefore, you
need more equations. Substituting t 5 t1 into the remaining projectile
motion equations gives
(2)
d 5 (v0 cos 45°) t
1 m 5 (v0 sin 45°) t1 2
1 2
gt 1
2
(3)
You now have three equations (1), (2), and (3) and three unknowns
t1, v0, and d. Using g 5 9.81 m/s2 and solving these three equations
give t1 5 0.3083 s and
v0 5 6.73 m/s
b
d 5 1.466 m b
REFLECT and THINK: All of these projectile problems are similar. You write down the governing equations for motion in the horizontal and vertical directions and then use additional information in
the problem statement to solve the problem. In this case, the distance
is just less than 1.5 meters, which is a reasonable distance for a
worker to toss a package.
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11.4
673
Curvilinear Motion of Particles
Sample Problem 11.13
Airplane B, which is travelling at a constant 560 km/h, is pursuing airplane
A, which is travelling northeast at a constant 800 km/hr. At time t 5 0,
airplane A is 640 km east of airplane B. Determine (a) the direction of
the course airplane B should follow (measured from the east) to intercept
plane A, (b) the rate at which the distance between the airplanes is
decreasing, (c) how long it takes for airplane B to catch airplane A.
STRATEGY: To find when B intercepts A, you just need to find out
when the two planes are at the same location. The rate at which the
distance is decreasing is the magnitude of vB/A, so you can use the relative
velocity equation.
MODELING and ANALYSIS: Choose x to be east, y to be north, and
place the origin of your coordinate system at B (Fig. 1).
y
vB
O
q
B
vA
45°
A
Positions of the Planes: You know that each plane has a constant
speed, so you can write a position vector for each plane. Thus,
x
rA 5 [(vA cos 45°) t 1 640 km]i 1 [(vA sin 45°) t] j
rB 5 [(vB cos θ) t]i 1 [(vB sin θ )t]j
640 km
Fig. 1 Initial velocity of
airplanes A and B.
(1)
(2)
a. Direction of B. Plane B will catch up when they are at the same
location, that is, rA 5 rB. You can equate components in the j direction to find
vA sin 45°t1 5 vB sin θ t1
After you substitute in values,
(vA sin 45°)t1
(560 km/hr)sin 45°
5
5 0.4950
vB t1
800 km/hr
θ 5 29.7º b
θ 5 sin21 0.4950 5 29.67°
sin θ 5
b. Rate. The rate at which the distance is decreasing is the magnitude
of vB/A, so
vB/A 5 vB 2 vA 5 (vB cos θ i 1 vB sin θ j) 2 (vA cos 45° i 1 vA sin 45° j)
5 [(800 km/h)cos 29.668° 2 (560 km/h)cos 45°]i
1 [(800 km/h)sin 29.668° 2(560 km/h)sin 45°]j
5 299.15 km/h i
ZvB/AZ 5 299 km/h b
c. Time for B to catch up with A. To find the time, you equate
the i components of each position vector, giving
(vA cos 45°) t1 1 640 km 5 (vB cos θ) t1
Solve this for t1. Thus,
t1 5
5
640 km
vB cos θ 2 vA cos 45°
640 km
5 2.139 h
(800 km/h)cos 29.67° 2 (560 km/h)cos 45°
t1 = 2.14 h b
REFLECT and THINK: The relative velocity is only in the horizontal
(eastern) direction. This makes sense, because the vertical (northern)
components have to be equal in order for the two planes to intersect.
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674
Kinematics of Particles
Sample Problem 11.14
Automobile A is traveling east at the constant speed of 36 km/h. As
automobile A crosses the intersection shown, automobile B starts from rest
35 m north of the intersection and moves south with a constant acceleration
of 1.2 m/s2. Determine the position, velocity, and acceleration of B relative
to A 5 s after A crosses the intersection.
B
1.2 m /s2
35 m
A
STRATEGY: This is a relative motion problem. Determine the motion
of each vehicle independently, and then use the definition of relative
motion to determine the desired quantities.
36 km /h
MODELING and ANALYSIS:
Motion of Automobile A. Choose x and y axes with the origin at
the intersection of the two streets and with positive senses directed east
and north, respectively. First express the speed in m/s, as
y
35 m
vA 5 a36
B
yB
A
x
xA
km 1000 m
1h
ba
ba
b 5 10 m/s
h
1 km
3600 s
The motion of A is uniform, so for any time t
aA 5 0
vA 5 110 m/s
xA 5 (xA)0 1 vAt 5 0 1 10t
Fig. 1 Initial positions of
car A and B.
For t 5 5 s, you have (Fig. 1)
aA 5 0
vA 5 10 m/s y
rA 5 50 m y
aA 5 0
vA 5 110 m/s
xA 5 1(10 m/s)(5 s) 5 150 m
Motion of Automobile B.
The motion of B is uniformly acceler-
ated, so
aB 5 21.2 m/s2
vB 5 (vB)0 1 at 5 0 2 1.2t
yB 5 (yB)0 1 (vB)0 t 1 12 aBt2 5 35 1 0 2 12(1.2)t2
For t 5 5 s, you have (Fig. 1)
rB/A
rB
vB
20 m
rB/A
a
rA
50 m
vA
10 m/s
vB/A
6 m/s
b
vB/A
aB 5 21.2 m/s2
vB 5 2(1.2 m/s2)(5 s) 5 26 m/s
yB 5 35 2 12(1.2 m/s2)(5 s)2 5 120 m
Motion of B Relative to A. Draw the triangle corresponding to the
vector equation rB 5 rA 1 rB/A (Fig. 2) and obtain the magnitude and
direction of the position vector of B relative to A.
rB/A 5 53.9 m
aB
aB/A
1.2 m/s2
aB/A
Fig. 2 Vector triangles for position,
velocity, and acceleration.
bee87342_ch11_615-717.indd 674
aB 5 1.2 m/s2w
vB 5 6 m/sw
rB 5 20 mx
α 5 21.8°
rB/A 5 53.9 m b 21.8° b
Proceeding in a similar fashion (Fig. 2), find the velocity and acceleration
of B relative to A. Hence,
vB/A 5 11.66 m/s
aB 5 aA 1 aB/A
vB 5 vA 1 vB/A
β 5 31.0°
vB/A 5 11.66 m/s d 31.0° b
aB/A 5 1.2 m/s2w b
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11.4
675
Curvilinear Motion of Particles
REFLECT and THINK: Note that the relative position and velocity of
B relative to A change with time; the values given here are only for the
moment t 5 5 s. Rather than drawing triangles, you could have also used
vector algebra. When the vectors are at right angles, as in this problem,
drawing vector triangles is usually easiest.
Sample Problem 11.15
Knowing that at the instant shown cylinder/ramp A has a velocity of
200 mm/s directed down, determine the velocity of block B.
STRATEGY: You have objects connected by cables, so this is a
dependent-motion problem. You should define coordinates for each blockobject and write a constraint equation for the cable. You will also need to
use relative motion, since B slides on A.
A
MODELING and ANALYSIS: Define position vectors, as shown in
q = 50°
Fig. 1.
B
Constraint Equations. Assuming the cable is inextensible, you can
write the length in terms of the coordinates and then differentiate.
The constraint equation for the cable is
xA 1 2xB/A 5 constant
Differentiating this gives
(1)
vA 5 22vB/A
xA
Substituting for vA gives vB/A 5 2100 mm/s or 100 mm/s up the incline.
A
xB/A
θ = 50
B
Dependent Motion: You know that the direction of vB/A is directed
up the incline. Therefore, the relative motion equation relating the velocities
of blocks A and B is vB 5 vA 1 vB/A. You could either draw a vector
triangle or use vector algebra. Let’s use vector algebra. Using the coordinate system shown in Fig. 2 and substituting in the magnitudes gives
(vB ) x i 1 (vB ) y j 5 (2200 mm/s)j 1 (2100 mm/s) sin 50° i
1 (100 mm/s) cos 50° j
Equating components gives
Fig. 1 Position vectors to A and B.
j
Coordinates for
vector algebra.
vB
vA = 200 mm/s
vB/A = 100 mm/s
40°
Fig. 3 Vector triangle for
velocity of blocks A and B.
bee87342_ch11_615-717.indd 675
j: (vB ) y 5 (2200 mm/s) 1 (100 mm/s)cos 50°
y vBy 5 2135.7 mm/s
vB 5 155.8 mm/s d 60.6° b
b
Fig. 2
y vBx 5 276.6 mm/s
Finding the magnitude and direction gives
a
i
i: (vB ) x 5 2(100 mm/s)sin 50°
REFLECT and THINK: Rather than using vector algebra, you could
have also drawn a vector triangle, as shown in Fig. 3. To use this vector
triangle, you need to use the law of cosines and the law of sines. Looking
at the mechanism, block B should move up the incline if block A moves
downward; our mathematical result is consistent with this. It is also
interesting to note that, even though B moves up the incline relative to A,
block B is actually moving down and to the left, as shown in the calculation
here. This occurs because block A is also moving down.
12/10/15 11:38 AM
SOLVING PROBLEMS
ON YOUR OWN
I
n the problems for this section, you will analyze the curvilinear motion of a
particle. The physical interpretations of velocity and acceleration are the same as
in the first sections of the chapter, but you should remember that these quantities are
vectors. In addition, recall from your experience with vectors in statics that it is often
advantageous to express position vectors, velocities, and accelerations in terms of their
rectangular scalar components [Eqs. (11.25) through (11.27)].
A. Analyzing the motion of a projectile. Many of the following problems deal with
the two-dimensional motion of a projectile where we can neglect the resistance of the
air. In Sec. 11.4C, we developed the equations that describe this type of motion, and
we observed that the horizontal component of the velocity remains constant (uniform
motion), while the vertical component of the acceleration is constant (uniformly accelerated motion). We are able to consider the horizontal and the vertical motions of the
particle separately. Assuming that the projectile is fired from the origin, we can write
the two equations as
x 5 (v x ) 0 t
y 5 (v y ) 0 t 2 12 gt 2
1. If you know the initial velocity and firing angle, you can obtain the value of y
corresponding to any given value of x (or the value of x for any value of y) by solving
one of the previous equations for t and substituting for t into the other equation
[Sample Prob. 11.10].
2. If you know the initial velocity and the coordinates of a point of the trajectory
and you wish to determine the firing angle α, begin your solution by expressing the
components (vx)0 and (vy)0 of the initial velocity as functions of α. Then substitute
these expressions and the known values of x and y into the previous equations. Finally,
solve the first equation for t and substitute that value of t into the second equation to
obtain a trigonometric equation in α, which you can solve for that unknown [Sample
Prob. 11.11].
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B. Solving translational two-dimensional relative-motion problems. You saw in
Sec. 11.4D that you can obtain the absolute motion of a particle B by combining the
motion of a particle A and the relative motion of B with respect to a frame attached
to A that is in translation [Sample Probs. 11.12 and 11.13]. You can then express the
velocity and acceleration of B as shown in Eqs. (11.32) and (11.33), respectively.
1. To visualize the relative motion of B with respect to A, imagine that you are
attached to particle A as you observe the motion of particle B. For example, to a
passenger in automobile A of Sample Prob. 11.14, automobile B appears to be heading
in a southwesterly direction (south should be obvious; west is due to the fact that
automobile A is moving to the east—automobile B then appears to travel to the west).
Note that this conclusion is consistent with the direction of vB/A.
2. To solve a relative-motion problem, first write the vector equations (11.30),
(11.32), and (11.33), which relate the motions of particles A and B. You may then use
either of the following methods.
a. Construct the corresponding vector triangles and solve them for the
desired position vector, velocity, and acceleration [Sample Prob. 11.14].
b. Express all vectors in terms of their rectangular components and solve
the resulting two independent sets of scalar equations [Sample Prob. 11.15]. If you
choose this approach, be sure to select the same positive direction for the displacement, velocity, and acceleration of each particle.
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Problems
CONCEPT QUESTIONS
11.CQ3 Two model rockets are fired simultaneously from a ledge and follow
the trajectories shown. Neglecting air resistance, which of the rockets will hit the ground first?
a. A.
b. B.
c. They hit at the same time.
d. The answer depends on h.
A
B
h
Fig. P6.CQ3
11.CQ4 Ball A is thrown straight up. Which of the following statements
about the ball are true at the highest point in its path?
a. The velocity and acceleration are both zero.
b. The velocity is zero, but the acceleration is not zero.
c. The velocity is not zero, but the acceleration is zero.
d. Neither the velocity nor the acceleration is zero.
h
v0
A
y
11.CQ5 Ball A is thrown straight up with an initial speed v0 and reaches a
Fig. P6.CQ4
(a)
(b)
(c)
maximum elevation h before falling back down. When A reaches its
maximum elevation, a second ball is thrown straight upward with
the same initial speed v0. At what height, y, will the balls cross
paths?
a. y 5 h
b. y . h/2
c. y 5 h/2
d. y , h/2
e. y 5(d)
0
(e)
11.CQ6 Two cars are approaching an intersection at constant speeds as shown.
What velocity will car B appear to have to an observer in car A?
a. S b. R c. a d. Q e. b
vB
vA
Fig. P6.CQ6
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B
11.CQ7 Blocks A and B are released from rest in the positions shown.
Neglecting friction between all surfaces, which figure best indicates
the direction α of the acceleration of block B?
b.
aB
c.
a = qa.
aB
d.
aB
a > qaB
b.
aa<=qq
a
aB B
e.c.
aB
a>q
d.
a<q
aB
e.
aB
B
q
B
q
A
A
Fig. P6.CQ7
END-OF-SECTION PROBLEMS
11.89 A ball is thrown so that the motion is defined by the equations
x 5 5t and y 5 2 1 6t 2 4.9t2, where x and y are expressed in meters
and t is expressed in seconds. Determine (a) the velocity at t 5 l s,
(b) the horizontal distance the ball travels before hitting the ground.
y
3
2
1
0
2
4
6
8
10 x
−1
y
−2
x
Fig. P11.90
Fig. P11.89
y
11.90 The motion of a vibrating particle is defined by the position vector
r 5 10(1 2 e23t)i 1 (4e22t sin 15t)j, where r and t are expressed
in millimeters and seconds, respectively. Determine the velocity
and acceleration when (a) t 5 0, (b) t 5 0.5 s.
11.91 The motion of a vibrating particle is defined by the position vector
r 5 (4 sin πt)i 2 (cos 2πt)j, where r is expressed in meters and
t in seconds. (a) Determine the velocity and acceleration when
t 5 1 s. (b) Show that the path of the particle is parabolic.
4m
4m
1m
O
1m x
Fig. P11.91
y/y1
11.92 The motion of a particle is defined by the equations x 5 100t 2
50 sin t and y 5 100 2 50 cos t, where x and y are expressed in
mm and t is expressed in seconds. Sketch the path of the particle
for the time interval 0 # t # 2π, and determine (a) the magnitudes
of the smallest and largest velocities reached by the particle, (b) the
corresponding times, positions, and directions of the velocities.
11.93 The damped motion of a vibrating particle is defined by the posi-
tion vector r 5 x1[1 2 1/(t 1 1)]i 1 (y1e2πt/2 cos 2πt)j, where t is
expressed in seconds. For x1 5 30 mm and y1 5 20 mm, determine
the position, the velocity, and the acceleration of the particle when
(a) t 5 0, (b) t 5 1.5 s.
1.0
0.5
0
0.2
0.4
0.6
x/x1
–0.5
–1.0
Fig. P11.93
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11.94 A girl operates a radio-controlled model car in a vacant parking lot.
The girl’s position is at the origin of the xy coordinate axes, and the
surface of the parking lot lies in the x–y plane. The motion of the car
is defined by the position vector r 5 (2 1 2t2)i 1 (6 1 t3) j where r and
t are expressed in meters and seconds, respectively. Determine (a)
the distance between the car and the girl when t 5 2 s, (b) the distance the car traveled in the interval from t 5 0 to t 5 2 s, (c) the
speed and direction of the car’s velocity at t 5 2 s, (d) the magnitude
of the car’s acceleration at t 5 2 s.
y (m)
6
0
x (m)
2
Fig. P11.94
11.95 The three-dimensional motion of a particle is defined by the position
y
2
2
y2
– x – z =1
A2 B2
A2
vector r 5 (Rt cos vnt)i 1 ctj 1 (Rt sin vnt)k. Determine the
magnitudes of the velocity and acceleration of the particle. (The
space curve described by the particle is a conic helix.)
*11.96 The three-dimensional motion of a particle is defined by the
z
Fig. P11.96
x
position vector r 5 (At cos t)i 1 (A 2t 2 1 1)j 1 (Bt sin t)k,
where r and t are expressed in meters and seconds, respectively.
Show that the curve described by the particle lies on the hyperboloid (y/A)2 2 (x/A)2 2 (z/B)2 5 1. For A 5 3 and B 5 1, determine
(a) the magnitudes of the velocity and acceleration when t 5 0,
(b) the smallest nonzero value of t for which the position vector and
the velocity are perpendicular to each other.
11.97 An airplane used to drop water on brushfires is flying horizontally
in a straight line at 315 km/h at an altitude of 80 m. Determine the
distance d at which the pilot should release the water so that it will
hit the fire at B.
v0
A
B
d
Fig. P11.97
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11.98 A ski jumper starts with a horizontal take-off velocity of 25 m/s and
lands on a straight landing hill inclined at 30°. Determine (a) the time
between take-off and landing, (b) the length d of the jump, (c) the
maximum vertical distance between the jumper and the landing hill.
25 m/s
d
30°
Fig. P11.98
11.99 A baseball pitching machine “throws” baseballs with a horizontal
velocity v0. Knowing that height h varies between 788 mm and
1068 mm, determine (a) the range of values of v0, (b) the values of
α corresponding to h 5 788 mm and h 5 1068 mm.
12.2 m
v0
A
α
1.5 m
B
h
Fig. P11.99
11.100 While delivering newspapers, a girl throws a newspaper with a
horizontal velocity v0. Determine the range of values of v0 if the
newspaper is to land between points B and C.
350 mm
900 mm
v0
A
1.2 m
C
200 mm
200 mm
200 mm
B
2.1 m
Fig. P11.100
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11.101 Water flows from a drain spout with an initial velocity of 0.75 m/s at
an angle of 15° with the horizontal. Determine the range of values of
the distance d for which the water will enter the trough BC.
A
15°
v0
3m
B
C
0.36 m
d
0.6 m
Fig. P11.101
11.102 In slow pitch softball, the underhand pitch must reach a maximum
height of between 1.8 m and 3.7 m above the ground. A pitch is
made with an initial velocity v0 with a magnitude of 13 m/s at an
angle of 33° with the horizontal. Determine (a) if the pitch meets the
maximum height requirement, (b) the height of the ball as it reaches
the batter.
v0
33°
0.6 m
15.2 m
Fig. P11.102
11.103 A volleyball player serves the ball with an initial velocity v0 of
magnitude 13.40 m/s at an angle of 20° with the horizontal.
Determine (a) if the ball will clear the top of the net, (b) how far
from the net the ball will land.
v0
A
C
20°
2.43 m
2.1 m
9m
Fig. P11.103
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11.104 A golfer hits a golf ball with an initial velocity of 50 m/s at an angle
of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between
the golfer and point B where the ball first lands.
v0
A
25°
5°
B
d
Fig. P11.104
11.105 A homeowner uses a snowblower to clear his driveway. Knowing
that the snow is discharged at an average angle of 40° with the
horizontal, determine the initial velocity v0 of the snow.
v0
B
40°
A
1.05 m
0.6 m
4.2 m
Fig. P11.105
11.106 At halftime of a football game souvenir balls are thrown to the
spectators with a velocity v0. Determine the range of values of v0 if
the balls are to land between points B and C.
7m
m
10
v0
A
40°
C
B
5m
d
35°
2 m 1.5 m
B
v0
8m
30°
Fig. P11.106
A
3m
2.1 m
11.107 A basketball player shoots when she is 5 m from the backboard.
Knowing that the ball has an initial velocity v0 at an angle of 30°
with the horizontal, determine the value of v0 when d is equal to
(a) 225 mm, (b) 425 mm.
Fig. P11.107
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11.108 A tennis player serves the ball at a height h 5 2.5 m with an initial
velocity of v0 at an angle of 5° with the horizontal. Determine the
range of v0 for which the ball will land in the service area that
extends to 6.4 m beyond the net.
5°
v0
h
0.914 m
12.2 m
6.4 m
Fig. P11.108
11.109 The nozzle at A discharges cooling water with an initial velocity v0 at
an angle of 6° with the horizontal onto a grinding wheel 350 mm in
diameter. Determine the range of values of the initial velocity for which
the water will land on the grinding wheel between points B and C.
20 mm
10°
C
6°
B
0.9 m
A
B
v0
C
30°
205 mm
5.7 m
200 mm
v0
65°
Fig. P11.109
A
0.7 m
5m
11.110 While holding one of its ends, a worker lobs a coil of rope over the
lowest limb of a tree. If he throws the rope with an initial velocity
v0 at an angle of 65° with the horizontal, determine the range of
values of v 0 for which the rope will go over only the lowest limb.
Fig. P11.110
11.111 The pitcher in a softball game throws a ball with an initial velocity
v0 of 72 km/h at an angle α with the horizontal. If the height of the
ball at point B is 0.68 m, determine (a) the angle α, (b) the angle θ
that the velocity of the ball at point B forms with the horizontal.
A
0.6 m
a
B
v0
0.68 m
q
vB
14 m
Fig. P11.111
684
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11.112 A model rocket is launched from point A with an initial velocity
v0 of 75 m/s. If the rocket’s descent parachute does not deploy and
the rocket lands a distance d 5 100 m from A, determine (a) the
angle α that v0 forms with the vertical, (b) the maximum height
above point A reached by the rocket, (c) the duration of the flight.
11.113 The initial velocity v0 of a hockey puck is 160 km/h. Determine
(a) the largest value (less than 45°) of the angle α for which the
puck will enter the net, (b) the corresponding time required for the
puck to reach the net.
C
D
v0
a
30°
v0
A
1.2 m
a
A
B
5m
d
E
B
0.75 m
Fig. P11.113
Fig. P11.112
11.114 A worker uses high-pressure water to clean the inside of a long
drainpipe. If the water is discharged with an initial velocity v0 of
11.5 m/s, determine (a) the distance d to the farthest point B on the
top of the pipe that the worker can wash from his position at A,
(b) the corresponding angle α.
d
B
v0
A
1.1 m
α
C
Fig. P11.114
11.115 An oscillating garden sprinkler which discharges water with an
initial velocity v0 of 8 m/s is used to water a vegetable garden.
Determine the distance d to the farthest point B that will be watered
and the corresponding angle α when (a) the vegetables are just
beginning to grow, (b) the height h of the corn is 1.8 m.
v0
h
a
A
B
1.5 m
d
Fig. P11.115
685
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*11.116 A ball is dropped onto a step at point A and rebounds with a veloc-
ity v0 at an angle of 15° with the vertical. Determine the value of
v0 knowing that just before the ball bounces at point B its velocity
vB forms an angle of 12° with the vertical.
v0
15˚
A
0.2 m
12˚
vB
B
Fig. P11.116
11.117 The velocities of skiers A and B are as shown. Determine the
velocity of A with respect to B.
14 m/s
B
D
25°
10 m/s
A
10°
A
B
C
Fig. P11.117
Fig. P11.118
11.118 The three blocks shown move with constant velocities. Find the
N
70°
S
A
90 km/h
velocity of each block, knowing that the relative velocity of A with
respect to C is 300 mm/s upward and that the relative velocity of B
with respect to A is 200 mm/s downward.
11.119 Three seconds after automobile B passes through the intersection
B
60 km/h
shown, automobile A passes through the same intersection.
Knowing that the speed of each automobile is constant, determine
(a) the relative velocity of B with respect to A, (b) the change
in position of B with respect to A during a 4-s interval, (c) the
distance between the two automobiles 2 s after A has passed
through the intersection.
Fig. P11.119
686
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11.120 Shore-based radar indicates that a ferry leaves its slip with a velocity
v 5 18 km/h d 70°, while instruments aboard the ferry indicate a
speed of 18.4 km/h and a heading of 30° west of south relative to
the river. Determine the velocity of the river.
11.121 Airplanes A and B are flying at the same altitude and are tracking
the eye of hurricane C. The relative velocity of C with respect to
A is vC/A 5 350 km/h d 75°, and the relative velocity of C with
respect to B is vC/B 5 400 km/h c 40°. Determine (a) the relative
velocity of B with respect to A, (b) the velocity of A if ground-based
radar indicates that the hurricane is moving at a speed of 30 km/h
due north, (c) the change in position of C with respect to B during
a 15-min interval.
Fig. P11.120
N
C
N
A
60°
70°
B
rse
cou
ding
hea
Fig. P11.121
11.122 Instruments in an airplane which is in level flight indicate that the
velocity relative to the air (airspeed) is 120 km/h and the direction
of the relative velocity vector (heading) is 70° east of north. Instruments on the ground indicate that the velocity of the airplane (ground
speed) is 110 km/h and the direction of flight (course) is 60° east of
north. Determine the wind speed and direction.
Fig. P11.122
11.123 Knowing that the velocity of block B with respect to block A is
vB/A 5 5.6 m/s a 70°, determine the velocities of A and B.
B
A
30°
qB
Fig. P11.123
687
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12/10/15 11:38 AM
11.124 Knowing that at the instant shown block A has a velocity of 200 mm/s
and an acceleration of 150 mm/s2 both directed down the incline,
determine (a) the velocity of block B, (b) the acceleration of block B.
11.125 A boat is moving to the right with a constant deceleration of
0.3 m/s2 when a boy standing on the deck D throws a ball with an
initial velocity relative to the deck which is vertical. The ball rises
to a maximum height of 8 m above the release point and the boy
must step forward a distance d to catch it at the same height as the
release point. Determine (a) the distance d, (b) the relative velocity
of the ball with respect to the deck when the ball is caught.
B
15°
25°
A
Fig. P11.124
8m
d
D
C
vD
aD = 0.3 m/s2
20°
Fig. P11.125
A
75°
B
11.126 The assembly of rod A and wedge B starts from rest and moves to the
right with a constant acceleration of 2 mm/s2. Determine (a) the
acceleration of wedge C, (b) the velocity of wedge C when t 5 10 s.
Fig. P11.126
11.127 Determine the required velocity of the belt B if the relative velocity with
which the sand hits belt B is to be (a) vertical, (b) as small as possible.
11.128 Conveyor belt A, which forms a 20° angle with the horizontal, moves
vA = 2.5 m/s
vB
B
1.5 m
at a constant speed of 1.2 m/s and is used to load an airplane.
Knowing that a worker tosses duffel bag B with an initial velocity
of 0.75 m/s at an angle of 30° with the horizontal, determine the
velocity of the bag relative to the belt as it lands on the belt.
A
15°
vA
Fig. P11.127
B
(vB)0
A
20°
30°
450 mm
Fig. P11.128
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11.129 During a rainstorm the paths of the raindrops appear to form an
N
angle of 30° with the vertical and to be directed to the left when
observed from a side window of a train moving at a speed of
15 km/h. A short time later, after the speed of the train has increased
to 24 km/h, the angle between the vertical and the paths of the drops
appears to be 45°. If the train were stopped, at what angle and with
what velocity would the drops be observed to fall?
A
30°
11.130 Instruments in airplane A indicate that, with respect to the air, the
plane is headed 30° north of east with an air speed of 480 km/h.
At the same time, radar on ship B indicates that the relative velocity
of the plane with respect to the ship is 416 km/h in the direction
33° north of east. Knowing that the ship is steaming due south at
20 km/h, determine (a) the velocity of the airplane, (b) the wind
speed and direction.
20 km/h
B
Fig. P11.130
11.131 When a small boat travels north at 5 km/h, a flag mounted on its stern
forms an angle θ 5 50° with the centerline of the boat as shown. A
short time later, when the boat travels east at 20 km/h, angle θ is again
50°. Determine the speed and the direction of the wind.
11.132 As part of a department store display, a model train D runs on a slight
incline between the store’s up and down escalators. When the train
and shoppers pass point A, the train appears to a shopper on the up
escalator B to move downward at an angle of 22° with the horizontal, and to a shopper on the down escalator C to move upward at an
angle of 23° with the horizontal and to travel to the left. Knowing
that the speed of the escalators is 1 m/s, determine the speed and
the direction of the train.
30°
q
Fig. P11.131
C
vC
D
A
vB
30°
B
Fig. P11.132
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690
Kinematics of Particles
11.5
NON-RECTANGULAR
COMPONENTS
Sometimes it is useful to analyze the motion of a particle in a coordinate
system that is not rectangular. In this section, we introduce two common
and important systems. The first system is based on the path of the particle;
the second system is based on the radial distance and angular displacement
of the particle.
11.5A Tangential and Normal
Components
We saw in Sec. 11.4 that the velocity of a particle is a vector tangent to
the path of the particle, but in general, the acceleration is not tangent to the
path. It is sometimes convenient to resolve the acceleration into components
directed, respectively, along the tangent and the normal to the path of the
particle. We will refer to this reference frame as tangential and normal
coordinates, which are sometimes called path coordinates.
y
Planar Motion of a Particle. First we consider a particle that
moves along a curve contained in a plane. Let P be the position of the
particle at a given instant. We attach at P a unit vector et tangent to the
path of the particle and pointing in the direction of motion (Fig. 11.19a).
Let e9t be the unit vector corresponding to the position P9 of the particle
at a later instant. Drawing both vectors from the same origin O9, we define
the vector Det 5 e9t 2 et (Fig. 11.19b). Since et and e9t are of unit length,
their tips lie on a circle with a radius of 1. Denote the angle formed by
et and e9t by Dθ. Then the magnitude of Det is 2 sin (Dθ/2). Considering
now the vector Det /Dθ, we note that, as Dθ approaches zero, this vector
becomes tangent to the unit circle of Fig. 11.19b, i.e., perpendicular to et,
and that its magnitude approaches
e't
en
P'
et
lim
Dθ y0
Thus, the vector obtained in the limit is a unit vector along the normal to
the path of the particle in the direction toward which et turns. Denoting
this vector by en, we have
P
O
2 sin(Dθ/2)
sin(Dθ/2)
51
5 lim
Dθ
Dθ y0
Dθ/2
x
(a)
e't
1
∆q
en 5 lim
Dθ y0
∆e t
en 5
et
O'
(b)
Fig. 11.19 (a) Unit tangent vectors for two
positions of particle P; (b) the angle between
the unit tangent vectors and their difference
Det.
bee87342_ch11_615-717.indd 690
det
dθ
Det
Dθ
(11.34)
Now, since the velocity v of the particle is tangent to the path, we
can express it as the product of the scalar v and the unit vector et. We
have
v 5 vet
(11.35)
12/10/15 11:38 AM
11.5
To obtain the acceleration of the particle, we differentiate Eq. (11.35) with
respect to t. Applying the rule for the differentiation of the product of a
scalar and a vector function (Sec. 11.4B), we have
a5
det
dv
dv
5 et 1v
dt
dt
dt
y
Non-Rectangular Components
∆s
∆q = ρ
e't
C
(11.36)
P'
ρ
However,
det
det dθ ds
5
dt
dθ ds dt
691
∆s
∆q
et
P
Recall from Eq. (11.15) that ds/dt 5 v, from Eq. (11.34) that det/dθ 5 en,
and from elementary calculus that dθ/ds is equal to 1/ρ, where ρ is the
radius of curvature of the path at P (Fig. 11.20). Then we have
det
v
5 en
r
dt
(11.37)
O
x
Fig. 11.20 Relationship among Dθ, Ds, and
ρ. Recall that for a circle, the arc length is
equal to the radius multiplied by the angle.
Substituting into Eq. (11.36), we obtain
Acceleration in normal
and tangential components
a5
v2
d
dv
et 1 en
r
dt
(11.38)
Thus, the scalar components of the acceleration are
at 5
dv
dt
y
C
2
an 5
v
r
These relations state that the tangential component of the acceleration is equal to the rate of change of the speed of the particle, whereas
the normal component is equal to the square of the speed divided by
the radius of curvature of the path at P. For a given speed, the normal
acceleration increases as the radius of curvature decreases. If the particle
travels in a straight line, then ρ is infinite, and the normal acceleration is
zero. If the speed of the particle increases, at is positive, and the vector
component at points in the direction of motion. If the speed of the particle
decreases, at is negative, and at points against the direction of motion. The
vector component an, on the other hand, is always directed toward the
center of curvature C of the path (Fig. 11.21).
We conclude from this discussion that the tangential component of
the acceleration reflects a change in the speed of the particle, whereas its
normal component reflects a change in the direction of motion of the
particle. The acceleration of a particle is zero only if both of its components are zero. Thus, the acceleration of a particle moving with constant
speed along a curve is not zero unless the particle happens to pass through
a point of inflection of the curve (where the radius of curvature is infinite)
or unless the curve is a straight line.
The fact that the normal component of acceleration depends upon
the radius of curvature of the particle’s path is taken into account in the
design of structures or mechanisms as widely different as airplane wings,
railroad tracks, and cams. In order to avoid sudden changes in the acceleration of the air particles flowing past a wing, wing profiles are designed
without any sudden change in curvature. Similar care is taken in designing
bee87342_ch11_615-717.indd 691
v2
a n = ρ en
(11.39)
at =
dv
et
dt
P
O
x
Fig. 11.21
Acceleration components in
normal and tangential coordinates; the
normal component always points toward the
center of curvature of the path.
12/10/15 11:38 AM
692
Kinematics of Particles
railroad curves to avoid sudden changes in the acceleration of the cars
(which would be hard on the equipment and unpleasant for the passengers).
A straight section of track, for instance, is never directly followed by a
circular section. Special transition sections are used to help pass smoothly
from the infinite radius of curvature of the straight section to the finite
radius of the circular track. Likewise, in the design of high-speed cams
(that can be used to transform rotary motion into translational motion),
abrupt changes in acceleration are avoided by using transition curves that
produce a continuous change in acceleration.
Motion of a Particle in Space. The relations in Eqs. (11.38) and
(11.39) still hold in the case of a particle moving along a space curve.
However, since an infinite number of straight lines are perpendicular to
the tangent at a given point P of a space curve, it is necessary to define
more precisely the direction of the unit vector en.
Let us consider again the unit vectors et and e9t tangent to the path
of the particle at two neighboring points P and P9 (Fig. 11.22a). Again
the vector Det represents the difference between et and e9t (Fig. 11.22b).
Let us now imagine a plane through P (Fig. 11.22c) parallel to the plane
defined by the vectors et, e9t, and Det (Fig. 11.22b). This plane contains
the tangent to the curve at P and is parallel to the tangent at P9. If we let
P9 approach P, we obtain in the limit the plane that fits the curve most
closely in the neighborhood of P. This plane is called the osculating plane
at P (from the Latin osculari, to kiss). It follows from this definition that
the osculating plane contains the unit vector en, since this vector represents
the limit of the vector Det /Dθ. The normal defined by en is thus contained
in the osculating plane; it is called the principal normal at P. The unit
vector eb 5 et 3 en that completes the right-handed triad et, en, and eb
(Fig. 11.22c) defines the binormal at P. The binormal is thus perpendicular
to the osculating plane. We conclude that the acceleration of the particle
at P can be resolved into two components: one along the tangent and the
other along the principal normal at P, as indicated in Eq. (11.38). Note
that the acceleration has no component along the binormal.
Photo 11.5 The passengers in a train
traveling around a curve experience a
normal acceleration toward the center of
curvature of the path.
y
y
eb
e't
P'
en
y'
et
P
et
P
e't
∆e t
∆θ
O
z
(a)
et
O'
x
z'
Osculating
plane
O
x'
(b)
z
x
(c)
Fig. 11.22 (a) Unit tangent vectors for a particle moving in space; (b) the plane defined by the unit vectors and the
vector difference Det; (c) the osculating plane contains the unit tangent and principal normal vectors and is
perpendicular to the unit binormal vector.
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11.5
11.5B
Non-Rectangular Components
693
Radial and Transverse
Components
In some situations in planar motion, the position of particle P is defined
by its polar coordinates r and θ (Fig. 11.23a). It is then convenient to
resolve the velocity and acceleration of the particle into components
parallel and perpendicular to the radial line OP. These components are
called radial and transverse components.
eθ
er
P
θ
∆eθ
P
r = re r
r
∆e r
er
∆θ
∆θ
O'
O
(a)
e'r
e'θ
θ
O
eθ
(c)
(b)
Fig. 11.23 (a) Polar coordinates r and θ of a particle at P; (b) radial and transverse unit vectors; (c) changes of
the radial and transverse unit vectors resulting from a change in angle Dθ.
We attach two unit vectors, er and eθ, at P (Fig. 11.23b). The vector
er is directed along OP and the vector eθ is obtained by rotating er through
90° counterclockwise. The unit vector er defines the radial direction, i.e.,
the direction in which P would move if r were increased and θ were kept
constant. The unit vector eθ defines the transverse direction, i.e., the
direction in which P would move if θ were increased and r were kept
constant. A derivation similar to the one we used in the preceding section
to determine the unit vector et leads to the relations
der
5 eθ
dθ
deθ
5 2er
dθ
(11.40)
Here 2er denotes a unit vector with a sense opposite to that of er
(Fig. 11.23c). Using the chain rule of differentiation, we express the time
derivatives of the unit vectors er and eθ as
der dθ
der
dθ
5
5 eθ
dt
dθ dt
dt
deθ dθ
deθ
dθ
5
5 2er
dt
dθ dt
dt
or using dots to indicate differentiation with respect to t as
.
.
er 5 θeθ
.
.
eθ 5 2θer
(11.41)
To obtain the velocity v of particle P, we express the position vector
r of P as the product of the scalar r and the unit vector er and then
differentiate with respect to t for
v5
bee87342_ch11_615-717.indd 693
d
.
.
(rer ) 5 rer 1 rer
dt
Photo 11.6 The foot pedals on an elliptical
trainer undergo curvilinear motion.
12/10/15 11:38 AM
694
Kinematics of Particles
Using the first of the relations of Eq. (11.41), we can rewrite this as
Velocity in radial and
transverse components
· 1 ru. e
v 5 re
r
u
(11.42)
Differentiating again with respect to t to obtain the acceleration, we have
a5
dv
5 r̈er 1 ṙ ėr 1 ṙθ̇eθ 1 rθ̈ eθ 1 rθ̇ ėθ
dt
Substituting for ėr and ėθ from Eq. (11.41) and factoring er and eθ, we
obtain
Acceleration in radial and
transverse components
.
..
a 5 (r̈ 2 rθ 2 )er 1 (rθ̈ 1 2r θ)eθ
(11.43)
The scalar components of the velocity and the acceleration in the radial
and transverse directions are
vr 5 ṙ
ar 5 r̈ 2 rθ̇
v 5 rθ̇eθ
P
z
O
y
R
x
(11.44)
aθ 5 rθ̈ 1 2 ṙθ̇
(11.45)
z
k
P
r
r 5 ReR 1 zk
eR
y
ReR
x
(b)
Fig. 11.24 (a) Cylindrical coordinates R, θ,
and z; (b) unit vectors in cylindrical
coordinates for a particle in space.
bee87342_ch11_615-717.indd 694
(11.46)
Compare this to using tangential and normal coordinates for a particle in
a circular path. In this case, the radius of curvature ρ is equal to the radius
of the circle r, and we have v 5 vet and a 5 v̇et 1(v2/r)en. Note that er
and en point in opposite directions (en inward and er outward).
eθ
zk
O
a 5 2rθ̇ 2er 1 rθ̈ eθ
Extension to the Motion of a Particle in Space: Cylindrical
Coordinates. Sometimes it is convenient to define the position of a
particle P in space by its cylindrical coordinates R, θ, and z (Fig. 11.24a).
We can then use the unit vectors eR, eθ, and k shown in Fig. 11.24b.
Resolving the position vector r of particle P into components along the
unit vectors, we have
(a)
θ
vθ 5 rθ̇
It is important to note that ar is not equal to the time derivative of vr and
that aθ is not equal to the time derivative of vθ.
In the case of a particle moving along a circle with a center O, we
have r 5 constant and ṙ 5 r̈ 5 0, so the formulas (11.42) and (11.43)
reduce, respectively, to
z
θ
2
(11.47)
Observe that eR and eθ define the radial and transverse directions in the
horizontal xy plane, respectively, and that the vector k, which defines the
axial direction, is constant in direction as well as in magnitude. Then we
can verify that
.
.
dr
.
5 ReR 1 Rθeθ 1 z k
dt
..
.
dv
a5
5 (R̈ 2 Rθ2 )eR 1 (Rθ̈ 1 2Rθ)eθ 1 z̈ k
dt
v5
(11.48)
(11.49)
12/10/15 11:38 AM
11.5
695
Non-Rectangular Components
Sample Problem 11.16
A motorist is traveling on a curved section of highway with a radius of
750 m at a speed of 90 km/h. The motorist suddenly applies the brakes,
causing the automobile to slow down at a constant rate. If the speed has
been reduced to 72 km/h after 8 s, determine the acceleration of the automobile immediately after the brakes have been applied.
vA = 90 km/h
A
750 m
STRATEGY: You know the path of the motion, and that the forward
speed of the vehicle defines the direction of et. Therefore, you can use
tangential and normal components.
MODELING and ANALYSIS:
Tangential Component of Acceleration.
First express the speeds
in m/s.
90 km / h 5 a90
1h
km 1000 m
ba
ba
b 5 25 m/s
h
1 km
3600 s
72 km/h 5 20 m/s
Since the automobile slows down at a constant rate, you have the tangential acceleration of
at 5 average at 5
20 m/s 2 25 m/s
Dv
5
5 20.625 m/s 2
Dt
8s
Normal Component of Acceleration. Immediately after the
brakes have been applied, the speed is still 88 ft/s. Therefore, you have
an 5
a t = 0.625 m/s 2
A
Motion
a n = 0.833 m/s 2
Fig. 1 Acceleration of
bee87342_ch11_615-717.indd 695
Magnitude and Direction of Acceleration. The magnitude and
direction of the resultant a of the components an and at are (Fig. 1)
a
a
the car.
(25 m/s) 2
v2
5 0.833 m/s2
5
r
750 m
tan α 5
a5
an
0.833 m/s2
5
at
0.625 m/s 2
an
0.833 m/s 2
5
sin α
sin 53.18
α 5 53.1° b
a 5 1.041 m/s 2
b
REFLECT and THINK: The tangential component of acceleration is
opposite the direction of motion, and the normal component of acceleration
points to the center of curvature, which is what you would expect for
slowing down on a curved path. Attempting to do this problem in Cartesian
coordinates is quite difficult.
12/10/15 11:38 AM
696
Kinematics of Particles
Sample Problem 11.17
Determine the minimum radius of curvature of the trajectory described by
the projectile considered in Sample Prob. 11.10.
STRATEGY: You are asked to find the radius of curvature, so you
should use normal and tangential coordinates.
v = vx
a = an
Fig. 1 Acceleration and velocity of
the projectile.
MODELING and ANALYSIS: Since an 5 v2/ρ, you have ρ 5 v2/an.
Therefore, the radius is small when v is small or when an is large. The
speed v is minimum at the top of the trajectory, since vy 5 0 at that point;
an is maximum at that same point, since the direction of the vertical coincides with the direction of the normal (Fig. 1). Therefore, the minimum
radius of curvature occurs at the top of the trajectory. At this point, you
have
v 5 vx 5 155.9 m/s
an 5 a 5 9.81 m/s2
2
(155.9 m/s) 2
v
r5
5
ρ 5 2480 m b
an
9.81 m/s2
REFLECT and THINK: The top of the trajectory is the easiest point to
determine the radius of curvature. At any other point in the trajectory, you
need to find the normal component of acceleration. You can do this easily
at the top, because you know that the total acceleration is pointed vertically downward and the normal component is simply the component
perpendicular to the tangent to the path. Once you have the normal
acceleration, it is straightforward to find the radius of curvature if you
know the speed.
Sample Problem 11.18
A
B
r
q
O
The rotation of the 0.9-m arm OA about O is defined by the relation
θ 5 0.15t2, where θ is expressed in radians and t in seconds. Collar B
slides along the arm in such a way that its distance from O is r 5 0.9
2 0.12t2, where r is expressed in meters and t in seconds. After the arm
OA has rotated through 30°, determine (a) the total velocity of the collar,
(b) the total acceleration of the collar, (c) the relative acceleration of the
collar with respect to the arm.
STRATEGY: You are given information in terms of r and θ, so you
should use polar coordinates.
MODELING and ANALYSIS: Model the collar as a particle.
Time t at which θ 5 30°.
Substitute θ 5 30° 5 0.524 rad into the
expression for θ. You obtain
θ 5 0.15t 2
bee87342_ch11_615-717.indd 696
0.524 5 0.15t 2
t 5 1.869 s
12/10/15 11:38 AM
11.5
eq
eq
B
eq
q
O
B
B
r
er
er
er
v = v re r + vqU eUq
a
eer +
aqU eand
q
U
vUq = (0.270 m /s)eUq
Fig.v1== avrrRadial
transverse
r + vqU e
U
q
coordinates
for
a = a re r + aqU eUqcollar B.
v
vUq = (0.270 m /s)eUq
B
b
vUq = (0.270 m /s)eUq
v
B
b m
v
1
B
8
vr = (–0.449
m /s)er
30° = 0.b4
O
r
m
81
30° = 0.4 m vr = (–0.449 m /s)er
O
r
81
vr = (–0.449 m /s)er
30° = 0.4
O
r
B
Fig.
2 Velocity
collar B.
a r = (–0.391
m/s2)eof
r
g
a r = (–0.391 m/s2)er g
ag
a
a B/OA = (–0.240
Equations of Motion. Substituting t 5 1.869 s in the expressions
for r, θ, and their first and second derivatives, you have
r 5 0.9 2 0.12t 2 5 0.481 m
θ 5 0.15t 2 5 0.524 rad
eq
ṙ 5 20.24t 5 20.449 m/s
θ̇ 5 0.30t 5 0.561 rad /s
er
2
2
θ̈ 5
r̈ 5 20.24 5 20.240 m/s
B 0.30 5 0.300 rad /s
a. Velocity of B. Using Eqs. (11.44), you can obtain the values of vr
and vθ when t 5 1.869 s (Fig. 1).
r
q
O v = v re r + vqU eUqr
q
O a = a re r + aqU eUq
a r = (–0.391 m/s2)er
697
Non-Rectangular Components
a
m/s2)er
B
B
aUq = (– 0.359 m /s 2)e q
q
O
r
vr 5 ṙ 5 20.449 m/s
vθ 5 rθ̇ 5 0.481(0.561) 5 0.270 m/s
v = v re r + vqU eUq
a = a re r + aqU eUq
Solve the right triangle shown in Fig. 2 to obtain the magnitude and
direction of the velocity,
vUq = (0.270 m /s)eUq
v 5 0.524 m/s
v
b. Acceleration of B.
B
Usingb Eqs. (11.45),
you obtain (Fig. 3)
ar 5 r̈ 2 rθ̇ 2
m
2
81 2v5
5 20.240 2 0.481(0.561)
20.391mm/s
/s)er
r = (–0.449
30° = 0.4
O
r
aθ 5 rθ̈ 1 2 ṙθ̇
5 0.481(0.300) 1 2(20.449)(0.561) 5 20.359 m/s2
a 5 0.531 m/s2
γ 5 42.6° b
c. Acceleration of B with Respect Bto Arm OA. Note that the
motion of the collar
respect
a r = with
(– 0.391
m /s2)erto the arm is rectilinear and defined by
the coordinate r (Fig. 4). You haveg
aB/OA 5 r̈ 5 20.240 m/s2
2
5 0.240
toward O. b
aUqa=B/OA
(– 0.359
m /s 2)em/s
q
aUq = (– 0.359Am /s 2)e q
aUq = (– 0.359 m /s 2)e q
B
A
a
Fig. 3 Acceleration of collar B. A
B
a B/OA = (–0.240 m/s2)er
O = (–0.240 m/s2)er
a B/OA
B
β 5 31.0° b
A
a B/OA = (–0.240 m /s2)er
B
O
O
O
Fig. 4
REFLECT and THINK: You should consider polar coordinates for any
kind of rotational motion. They turn this problem into a straightforward
solution, whereas any other coordinate system would make this problem
much more difficult. One way to make this problem harder would be to
ask you to find the radius of curvature in addition to the velocity and
acceleration. To do this, you would have to find the normal component of
the acceleration; that is, the component of acceleration that is perpendicular
to the tangential direction defined by the velocity vector.
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698
Kinematics of Particles
Sample Problem 11.19
6 km/h
75 m
r
60 m
v
A boy is flying a kite that is 60 m high with 75 m of cord out. The kite
moves horizontally from this position at a constant 6 km/h that is directly
away from the boy. Ignoring the sag in the cord, determine how fast the
cord is being let out at this instant and how fast this rate is increasing.
STRATEGY: The most natural way to describe the position of the kite
is using a radial vector and angle, as shown in Fig. 1. The distance r is
changing, so use polar coordinates.
eθ
er
v
75 m
r
60 m
q
O
Fig. 1 Radial and transverse
coordinates for the kite.
MODELING and ANALYSIS: The angle and the speed of the kite in
m/s are found by
θ 5 sin21a
hr
1000 m
5
60
km
b 5 53.13° and v 5 6 a ba
ba
b 5 m/s
75
hr 3600 s
km
3
Velocity in Polar Coordinates: You know that in polar coordinates
the velocity is v 5 ṙer 1 rθ̇ er. Using Fig. 1, you can resolve the velocity
vector into polar coordinates, giving
5
.
r 5 v cos θ 5 a m/sb cos 53.13°
3
.
rθ 5 2v sinθ
ṙ 5 1.000 m/s b
.
(5/3 m/s)sin 53.13°
v sinθ
5 0.01778 rad/s
θ52
52
r
75 m
Acceleration in Polar Coordinates: You know that the acceleration is zero, because the kite is traveling at a constant speed. This means
that both components of the acceleration need to be zero. You know the
radial component is ar 5 r̈ 2 rθ̇ 2 5 0. So
r̈ 5 rθ̇ 2 5 (75 m)(20.01778 rad/s)2
r̈ 5 0.0237 m/s2
b
REFLECT and THINK: When the angle is 90°, then ṙ will be zero.
When the angle is very small––that is, when the kite is far away––you
would expect the cord to increase at a rate of 6 m/s, which is the speed
of the kite. Our answer is reasonable since it is between these two
limits.
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11.5
699
Non-Rectangular Components
Sample Problem 11.20
6m
B
θ = 30
A
At the instant shown, the length of the boom AB is being decreased at the
constant rate of 0.2 m/s, and the boom is being lowered at the constant
rate of 0.08 rad/s. Determine (a) the velocity of point B, (b) the acceleration
of point B.
STRATEGY: Use polar coordinates, since that is the most natural way
to describe the position of point B.
MODELING and ANALYSIS: From the problem statement, you know
ṙ 5 20.2 m/s
θ̇ 5 20.08 rad /s
r̈ 5 0
θ̈ 5 0
a. Velocity of B. Using Eqs.(11.44), you can determine the values of
vr and vθ at this instant to be
vr 5 ṙ 5 20.2 m/s
vθ 5 rθ̇ 5 (6 m)(20.08 rad/s) 5 20.48 m/s
Therefore, you can write the velocity vector as
v 5 (20.200 m/s)er 1 (20.480 m/s)et b
b. Acceleration of B. Using Eqs. (11.45), you find
ar 5 r̈ 2 rθ̇ 2 5 0 2 (6 m)(20.08 rad/s)2 5 20.0384 m/s2
aθ 5 rθ̈ 1 2 ṙθ̇ 5 0 1 2(20.02 m/s)(20.08 rad/s) 5 0.00320 m/s2
or
eθ
er
30°
60°
a 5 (20.0384 m/s2)er 1 (0.00320 m/s2)eθ b
vr = –0.2 m/s B
vθ = –0.48 m/s
Fig. 1 Velocity of B.
B
REFLECT and THINK: Once you identify what you are given in the
problem statement, this problem is quite straightforward. Sometimes you
will be asked to express your answer in terms of a magnitude and direction.
The easiest way is to first determine the x and y components and then to
find the magnitude and direction. From Fig. 1,
1
y
: (vB)x 5 0.48 cos 60° 2 0.2 cos 30° 5 0.06680 m/s
1x:
vx
(vB)y 5 20.48 sin 60° 2 0.2 sin 30° 5 20.5157 m/s
So the magnitude and direction are
vB 5 20.066802 1 0.51572
vr
5 0.520 m/s
vθ
vy
vB
Fig. 2 Resultant velocity
of collar B in Cartesian
and in radial and
transverse coordinates.
bee87342_ch11_615-717.indd 699
tan β 5
0.51569
, β 5 82.6°
0.06680
So, an alternative way of expressing the velocity of B is vB 5 0.520 m/s c 82.6°
You could also find the magnitude and direction of the acceleration if you
needed it expressed in this way. It is important to note that no matter what
coordinate system we choose, the resultant velocity vector is the same.
You can choose to express this vector in whatever coordinate system is
most useful. Figure 2 shows the velocity vector vB resolved into x and y
components and r and θ coordinates.
12/10/15 11:38 AM
SOLVING PROBLEMS
ON YOUR OWN
I
n the following problems, you will be asked to express the velocity and the acceleration of particles in terms of either their tangential and normal components or
their radial and transverse components. Although these components may not be as
familiar to you as rectangular components, you will find that they can simplify the
solution of many problems and that certain types of motion are more easily described
when they are used.
1. Using tangential and normal components. These components are most often
used when the particle of interest travels along a known curvilinear path or when the
radius of curvature of the path is to be determined [Sample Prob. 11.16]. Remember
that the unit vector et is tangent to the path of the particle (and thus aligned with the
velocity), whereas the unit vector en is directed along the normal to the path and
always points toward its center of curvature. It follows that the directions of the two
unit vectors are constantly changing as the particle moves.
2. Acceleration in terms of tangential and normal components. We derived in
Sec. 11.5A the following equation, which is applicable to both the two-dimensional
and the three-dimensional motion of a particle:
a5
dv
v2
et 1 en
r
dt
(11.38)
The following observations may help you in solving the problems of this section.
a. The tangential component of the acceleration measures the rate of change
of the speed as at 5 dv/dt. It follows that, when at is constant, you can use the
equations for uniformly accelerated motion with the acceleration equal to at. Furthermore, when a particle moves at a constant speed, we have at 5 0, and the acceleration
of the particle reduces to its normal component.
b. The normal component of the acceleration is always directed toward the
center of curvature of the path of the particle, and its magnitude is an 5 v2/ρ. Thus,
you can determine the normal component if you know the speed of the particle and
the radius of curvature ρ of the path. Conversely, if you know the speed and normal
acceleration of the particle, you can find the radius of curvature of the path by solving
this equation for ρ [Sample Prob. 11.17].
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3. Using radial and transverse components. These components are used to analyze
the planar motion of a particle P when the position of P is defined by its polar
coordinates r and θ. As shown in Fig. 11.23, the unit vector er, which defines the
radial direction, is attached to P and points away from the fixed point O, whereas
the unit vector eθ, which defines the transverse direction, is obtained by rotating er
counterclockwise through 90°. The velocity and acceleration of a particle are expressed
in terms of their radial and transverse components in Eqs. (11.42) and (11.43), respectively. Note that the expressions obtained contain the first and second derivatives with
respect to t of both coordinates r and θ.
In the problems of this section, you will encounter the following types of problems
involving radial and transverse components.
a. Both r and θ are known functions of t. In this case, you compute the first
and second derivatives of r and θ and substitute the resulting expressions into
Eqs. (11.42) and (11.43).
b. A certain relationship exists between r and θ. First, you should determine
this relationship from the geometry of the given system and use it to express r as a
function of θ. Once you know the function r 5 f(θ), you can apply the chain rule to
determine ṙ in terms of θ and θ̇ , and r̈ in terms of θ, θ̇, and θ̈:
ṙ 5 f 9(θ)θ̇
r̈ 5 f 0(θ)θ̇ 2 1 f 9(θ)θ̈
You can then substitute these expressions into Eqs. (11.42) and (11.43).
c. The three-dimensional motion of a particle, as indicated at the end of
Sec. 11.5B, often can be described effectively in terms of the cylindrical coordinates
R, θ, and z (Fig. 11.24). The unit vectors then should consist of eR, eθ, and k. The
corresponding components of the velocity and the acceleration are given in Eqs. (11.48)
and (11.49). Note that the radial distance R is always measured in a plane parallel to
the xy plane, and be careful not to confuse the position vector r with its radial
component ReR.
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Problems
CONCEPT QUESTIONS
11.CQ8 The Ferris wheel is rotating with a constant angular velocity v.
What is the direction of the acceleration of point A?
a. y b. x c. w d. z e. The acceleration is zero.
11.CQ9 A race car travels around the track shown at a constant speed. At
which point will the race car have the largest acceleration?
a. A. b. B. c. C. d. D. e. The acceleration will be zero at all
the points.
C
A
v
B
Fig. P11.CQ8
A
D
Fig. P11.CQ9
u
11.CQ10 A child walks across merry-go-round A with a constant speed u
relative to A. The merry-go-round undergoes fixed-axis rotation
about its center with a constant angular velocity v counterclockwise.
When the child is at the center of A, as shown, what is the direction
of his acceleration when viewed from above?
a. y b. z c. x d. w e. The acceleration is zero.
A
ω
END-OF-SECTION PROBLEMS
11.133 Determine the smallest radius that should be used for a highway
if the normal component of the acceleration of a car traveling at
72 km/h is not to exceed 0.8 m/s2.
Fig. P11.CQ10
B
ρ
A
A
r
B
Fig. P11.133
11.134 Determine the maximum speed that the cars of the roller-coaster can
Fig. P11.134
reach along the circular portion AB of the track if ρ 5 25 m and the
normal component of their acceleration cannot exceed 3g.
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11.135 Human centrifuges are often used to simulate different acceleration
levels for pilots and astronauts. Space shuttle pilots typically face
inwards towards the center of the gondola in order to experience
a simulated 3-g forward acceleration. Knowing that the astronaut
sits 5 m from the axis of rotation and experiences 3 g’s inward,
determine her velocity.
5m
A
Fig. P11.135
11.136 Pin A, which is attached to link AB, is constrained to move in the
circular slot CD. Knowing that at t 5 0 the pin starts from rest and
moves so that its speed increases at a constant rate of 20 mm/s2,
determine the magnitude of its total acceleration when (a) t 5 0,
(b) t 5 2 s.
D
C
90 mm
A
B
Fig. P11.136
P
11.137 A monorail train starts from rest on a curve of radius 400 m and
accelerates at the constant rate at. If the maximum total acceleration of the train must not exceed 1.5 m/s2, determine (a) the shortest
distance in which the train can reach a speed of 72 km/h, (b) the
corresponding constant rate of acceleration at.
0.8 m
11.138 A robot arm moves so that P travels in a circle about point B, which is
not moving. Knowing that P starts from rest, and its speed increases
at a constant rate of 10 mm/s2, determine (a) the magnitude of the
acceleration when t 5 4 s, (b) the time for the magnitude of the
acceleration to be 80 mm/s2.
O
B
Fig. P11.138
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11.139 A monorail train starts from rest on a curve of radius 400 m and
B
accelerates at the constant rate at. If the maximum total acceleration
of the train must not exceed 1.5 m/s2, determine (a) the shortest
distance in which the train can reach a speed of 72 km/h, (b) the
corresponding constant rate of acceleration at.
C
100 m
11.140 A motorist starts from rest at point A on a circular entrance ramp
when t 5 0, increases the speed of her automobile at a constant rate
and enters the highway at point B. Knowing that her speed continues
to increase at the same rate until it reaches 100 km/h at point C,
determine (a) the speed at point B, (b) the magnitude of the total
acceleration when t 5 20 s.
150 m
A
Fig. P11.140
11.141 Race car A is traveling on a straight portion of the track while race
car B is traveling on a circular portion of the track. At the instant
shown, the speed of A is increasing at the rate of 10 m/s2, and the
speed of B is decreasing at the rate of 6 m/s2. For the position shown,
determine (a) the velocity of B relative to A, (b) the acceleration of
B relative to A.
240 km/h
A
B
50°
300 m
200 km/h
Fig. P11.141
11.142 At a given instant in an airplane race, airplane A is flying horizontally
in a straight line, and its speed is being increased at the rate of 8 m/s2.
Airplane B is flying at the same altitude as airplane A and, as it
rounds a pylon, is following a circular path of 300-m radius. Knowing
that at the given instant the speed of B is being decreased at the rate
of 3 m/s2, determine, for the positions shown, (a) the velocity of B
relative to A, (b) the acceleration of B relative to A.
400 m
A
450 km/h
B
300 m
120 km/h
y
30°
q
r = 70 m
540 km/h
P
x
Fig. P11.142
11.143 A race car enters the circular portion of a track that has a radius of
Fig. P11.143
70 m. When the car enters the curve at point P, it is travelling with
a speed of 120 km/h that is increasing at 5 m/s2. Three seconds later,
determine the x and y components of velocity and acceleration of
the car.
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11.144 An airplane flying at a constant speed of 240 m/s makes a banked
horizontal turn. What is the minimum allowable radius of the turn
if the structural specifications require that the acceleration of the
airplane shall never exceed 4 g?
11.145 A golfer hits a golf ball from point A with an initial velocity of
50 m/s at an angle of 25° with the horizontal. Determine the radius
of curvature of the trajectory described by the ball (a) at point A,
(b) at the highest point of the trajectory.
11.146 Three children are throwing snowballs at each other. Child A throws
a snowball with a horizontal velocity v0. If the snowball just passes
over the head of child B and hits child C, determine the radius of
curvature of the trajectory described by the snowball (a) at point B,
(b) at point C.
Fig. P11.144
vA
v0
A
A
1m
25°
B
C
7m
2m
Fig. P11.145
d
Fig. P11.146
11.147 Coal is discharged from the tailgate A of a dump truck with an
initial velocity vA 5 2 m/s d 50°. Determine the radius of curvature
of the trajectory described by the coal (a) at point A, (b) at the point
of the trajectory 1 m below point A.
A
50°
vA
Fig. P11.147
11.148 From measurements of a photograph, it has been found that as
the stream of water shown left the nozzle at A, it had a radius
of curvature of 25 m. Determine (a) the initial velocity vA of the
stream, (b) the radius of curvature of the stream as it reaches its
maximum height at B.
11.149 A child throws a ball from point A with an initial velocity v0 at an
angle of 3° with the horizontal. Knowing that the ball hits a wall at
point B, determine (a) the magnitude of the initial velocity, (b) the
minimum radius of curvature of the trajectory.
3°
B
vA
3
A
4
Fig. P11.148
v0
A
B
1.5 m
0.97 m
6m
Fig. P11.149
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11.150 A projectile is fired from point A with an initial velocity v0. (a) Show
that the radius of curvature of the trajectory of the projectile
reaches its minimum value at the highest point B of the trajectory.
(b) Denoting by θ the angle formed by the trajectory and the
horizontal at a given point C, show that the radius of curvature of
the trajectory at C is ρ 5 ρmin /cos3θ.
x
B
v0
C
ρ min
θ
α
A
ρ
Fig. P11.150
*11.151 Determine the radius of curvature of the path described by the
particle of Prob. 11.95 when t 5 0.
*11.152 Determine the radius of curvature of the path described by the
particle of Prob. 11.96 when t 5 0, A 5 3, and B 5 1.
11.153 and 11.154 A satellite will travel indefinitely in a circular orbit
around a planet if the normal component of the acceleration of the
satellite is equal to g(R/r)2, where g is the acceleration of gravity
at the surface of the planet, R is the radius of the planet, and r is
the distance from the center of the planet to the satellite. Knowing
that the diameter of the sun is 1.39 Gm and that the acceleration of
gravity at its surface is 274 m/s2, determine the radius of the orbit
of the indicated planet around the sun assuming that the orbit is
circular.
11.153 Earth: (y mean)orbit 5 107 Mm/h.
11.154 Saturn: (y mean)orbit 5 34.7 Mm/h.
11.155 through 11.157 Determine the speed of a satellite relative to the
indicated planet if the satellite is to travel indefinitely in a circular
orbit 160 km above the surface of the planet. (See information given
in Probs. 11.153–11.154.)
11.155 Venus: g 5 8.53 m/s2, R 5 6161 km.
11.156 Mars: g 5 3.83 m/s2, R 5 3332 km.
11.157 Jupiter: g 5 26.0 m/s2, R 5 69,893 km.
11.158 A satellite will travel indefinitely in a circular orbit around the earth
if the normal component of its acceleration is equal to g(R/r)2, where
g 5 9.81 m/s2, R 5 radius of the earth 5 6370 km, and r 5 distance
from the center of the earth to the satellite. Assuming that the orbit
of the moon is a circle with a radius of 384 3 103 km, determine
the speed of the moon relative to the earth.
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11.159 Knowing that the radius of the earth is 6370 km, determine the
B
time of one orbit of the Hubble Space Telescope if the telescope
travels in a circular orbit 590 km above the surface of the earth.
(See information given in Probs. 11.153–11.154.)
A
rB
11.160 Satellites A and B are traveling in the same plane in circular orbits
around the earth at altitudes of 180 and 300 km, respectively. If at
t 5 0 the satellites are aligned as shown and knowing that the radius
of the earth is R 5 6370 km, determine when the satellites will next
be radially aligned. (See information given in Probs. 11.153–11.154.)
11.161 The oscillation of rod OA about O is defined by the relation
θ 5 (2yπ)(sin πt), where θ and t are expressed in radians and
seconds, respectively. Collar B slides along the rod so that its
625
distance from O is r 5
where r and t are expressed in mm
(t + 4)
rA
Fig. P11.160
O
and seconds, respectively. When t 5 1 s, determine (a) the velocity
of the collar, (b) the acceleration of the collar, (c) the acceleration
of the collar relative to the rod.
θ
r
11.162 The path of a particle P is a limaçon. The motion of the particle is
B
defined by the relations r 5 b(2 1 cos πt) and θ 5 πt where t and θ
are expressed in seconds and radians, respectively. Determine (a) the
velocity and the acceleration of the particle when t 5 2 s, (b) the
value of θ for which the magnitude of the velocity is maximum.
11.163 During a parasailing ride, the boat is traveling at a constant 30 km/hr
with a 200-m long tow line. At the instant shown, the angle between
the line and the water is 30° and is increasing at a constant rate of
2°/s. Determine the velocity and acceleration of the parasailer at
this instant.
A
Fig. P11.161
P
r
q
Fig. P11.162
r
B
θ
v0
A
P
h
Fig. P11.163
b
O
11.164 Pin P is attached to BC and
· slides freely in the slot of OA. Determine the rate of change u of the angle u, knowing that BC moves
at a constant speed v0. Express your answer in terms of v0, h, b, Fig. P11.164
and u.
u
C
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D
A
r
θ
O
P
11.165 As rod OA rotates, pin P moves along the parabola BCD. Knowing
that the equation of this parabola is r 5 2b/(1 1 cos θ) and that
θ 5 kt, determine the velocity and acceleration of P when (a) θ 5 0,
(b) θ 5 90°.
11.166 The pin at B is free to slide along the circular slot DE and along
the rotating
rod OC. Assuming that the rod OC rotates at a constant
þ
rate u, (a) show that the acceleration of pin B is of constant magnitude,
(b) determine the direction of the acceleration of pin B.
C
b
B
D
r
Fig. P11.165
B
C
θ
b
O
B
A
v a
b
E
r
A
Fig. P11.166
θ
b
C
Fig. P11.167
11.167 To study the performance of a race car, a high-speed camera is
positioned at point A. The camera is mounted on a mechanism
which permits it to record the motion of the car as the car travels
on straightaway BC. Determine (a) the speed of the car in terms of
b, θ, and θ̇ , (b) the magnitude of the acceleration in terms of b, θ,
θ̇, and ü .
11.168 After taking off, a helicopter climbs in a straight line at a constant
angle β. Its flight is tracked by radar from point A. Determine the
speed of the helicopter in terms of d, β, θ, and θ̇.
v
B
A
β
θ
d
Fig. P11.168
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11.169 At the bottom of a loop in the vertical plane an airplane has a
horizontal velocity of 150 m/s and is speeding up at a rate of
25 m/s2. The radius of curvature of the loop is 2000 m. The plane
is being tracked by radar at O. What are the recorded values of ṙ,
r̈, θ̇, and θ̈ for this instant?
2000 m
150 m/s
600 m
r
q
O
800 m
Fig. P11.169
11.170 Pin C is attached to rod BC and slides freely in the slot of rod
OA which rotates at the constant
rate v. At the instant when
.
.
θ̈ . Express your answers
β 5 60°, determine (a) r and θ, (b) r̈ and θ
in terms of d and v.
A
C
r
O
q
B
b
d
d
Fig. P11.170
11.171 For the race car of Prob. 11.167, it was found that it took 0.5 s for
the car to travel from the position θ 5 60° to the position θ 5 35°.
Knowing that b 5 25 m, determine the average speed of the car
during the 0.5-s interval.
11.172 For the helicopter of Prob. 11.168, it was found that when the
helicopter was at B, the distance and the angle of elevation of the
helicopter were r 5 1000 m and θ 5 20°, respectively. Four seconds
later, the radar station sighted the helicopter at r 5 1100 m and
θ 5 23.1°. Determine the average speed and the angle of climb β
of the helicopter during the 4-s interval.
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11.173 and 11.174 A particle moves along the spiral shown. Determine
the magnitude of the velocity of the particle in terms of b, θ, and θ̇.
rq 2 = b
1 2
q
r = be 2
O
O
Fig. P11.173 and P11.175
Fig. P11.174 and P11.176
11.175 and 11.176 A particle moves along the spiral shown. Knowing
that θ̇ is constant and denoting this constant by v, determine the
magnitude of the acceleration of the particle in terms of b, θ, and θ̇.
11.177 The motion of a particle on the surface of a right circular cylinder is
defined by the relations R 5 A, θ 5 2πt, and z 5 B sin 2πnt, where
A and B are constants and n is an integer. Determine the magnitudes
of the velocity and acceleration of the particle at any time t.
z
A
B
B
x
.
.
11.178 Show that r 5 hϕ sin θ knowing that at the instant shown, step
AB ofþthe step exerciser is rotating counterclockwise at a constant
rate f.
h
θ r
B
P
d
A
Fig. P11.178
n = 10
Fig. P11.177
O
φ
y
11.179 The three-dimensional motion of a particle is defined by the
relations R 5 A(1 2 e2t), θ 5 2πt, and z 5 B(1 2 e2t ). Determine
the magnitudes of the velocity and acceleration when (a) t 5 0,
(b) t 5 `.
*11.180 For the conic helix of Prob. 11.95, determine the angle that the
osculating plane forms with the y axis.
*11.181 Determine the direction of the binormal of the path described by
the particle of Prob. 11.96 when (a) t 5 0, (b) t 5 π/2 s.
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Review and Summary
Position Coordinate of a Particle in Rectilinear Motion
In the first half of this chapter, we analyzed the rectilinear motion of a
particle, i.e., the motion of a particle along a straight line. To define the position P of the particle on that line, we chose a fixed origin O and a positive
direction (Fig. 11.25). The distance x from O to P, with the appropriate sign,
completely defines the position of the particle on the line and is called the
position coordinate of the particle [Sec. 11.1A].
O
P
x
x
Fig. 11.25
Velocity and Acceleration in Rectilinear Motion
The velocity v of the particle was shown to be equal to the time derivative of
the position coordinate x, so
v5
dx
dt
(11.1)
And we obtained the acceleration a by differentiating v with respect to t, as
a5
dv
dt
(11.2)
a5
d 2x
dt 2
(11.3)
or
We also noted that a could be expressed as
a5v
dv
dx
(11.4)
We observed that the velocity v and the acceleration a are represented
by algebraic numbers that can be positive or negative. A positive value for v
indicates that the particle moves in the positive direction, and a negative value
shows that it moves in the negative direction. A positive value for a, however,
may mean that the particle is truly accelerated (i.e., moves faster) in the
positive direction or that it is decelerated (i.e., moves more slowly) in the
negative direction. A negative value for a is subject to a similar interpretation
[Sample Prob. 11.1].
Determination of the Velocity and Acceleration by
Integration
In most problems, the conditions of motion of a particle are defined by the
type of acceleration that the particle possesses and by the initial conditions
[Sec. 11.1B]. Then we can obtain the velocity and position of the particle by
integrating two of the equations (11.1) to (11.4). The selection of these equations depends upon the type of acceleration involved [Sample Probs. 11.2
through 11.4].
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Uniform Rectilinear Motion
Two types of motion are frequently encountered. Uniform rectilinear motion
[Sec. 11.2A], in which the velocity v of the particle is constant, is described by
(11.5)
x 5 x0 1 vt
Uniformly Accelerated Rectilinear Motion
Uniformly accelerated rectilinear motion [Sec. 11.2B], in which the acceleration a of the particle is constant, is described by
(11.6)
v 5 v0 1 at
x 5 x 0 1 v 0t 1
1 2
2 at
v2 5 v20 1 2a(x 2 x0 )
A
O
xB/A
xB
Fig. 11.26
Relative Motion of Two Particles
B
xA
(11.7)
(11.8)
x
When two particles A and B (such as two aircraft) move, we may wish to
consider the relative motion of B with respect to A [Sec. 11.2C]. Denoting
the relative position coordinate of B with respect to A by xB/A (Fig. 11.26),
we have
xB 5 xA 1 xB/A
(11.9)
Differentiating Eq. (11.9) twice with respect to t, we obtained successively
vB 5 vA 1 vB/A
(11.10)
aB 5 aA 1 aB/A
(11.11)
where vB/A and aB/A represent, respectively, the relative velocity and the
relative acceleration of B with respect to A.
Dependent Motion
When several blocks are connected by inextensible cords, it is possible to
write a linear relation between their position coordinates. We can then write
similar relations between their velocities and between their accelerations,
which we can use to analyze their motion [Sample Probs. 11.7 and 11.8].
Graphical Solutions
It is sometimes convenient to use a graphical solution for problems involving
the rectilinear motion of a particle [Sec. 11.3]. The graphical solution most
commonly used involves the x–t, v–t, and a–t curves [Sample Prob. 11.10]. It
was shown at any given time t that
v 5 slope of x–t curve
a 5 slope of v–t curve
Also, over any given time interval from t1 to t2, we have
v2 2 v1 5 area under a–t curve
x2 2 x1 5 area under v–t curve
Position Vector and Velocity in Curvilinear Motion
In the second half of this chapter, we analyzed the curvilinear motion of a
particle, i.e., the motion of a particle along a curved path. We defined the
position P of the particle at a given time [Sec. 11.4A] by the position vector r
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joining the O of the coordinates and point P (Fig. 11.27). We defined the
velocity v of the particle by the relation
v5
dr
dt
ds
dt
v
(11.14)
The velocity is a vector tangent to the path of the particle with a magnitude
v (called the speed of the particle) equal to the time derivative of the length
s of the arc described by the particle. Thus,
v5
y
(11.15)
P
r
s
P0
O
x
Fig. 11.27
Acceleration in Curvilinear Motion
We defined the acceleration a of the particle by the relation
a5
dv
dt
(11.17)
and we noted that, in general, the acceleration is not tangent to the path of
the particle.
Derivative of a Vector Function
Before proceeding to the consideration of the components of velocity and
acceleration, we reviewed the formal definition of the derivative of a vector
function and established a few rules governing the differentiation of sums and
products of vector functions. We then showed that the rate of change of a
vector is the same with respect both to a fixed frame and to a frame in translation [Sec. 11.4B].
Rectangular Components of Velocity and Acceleration
Denoting the rectangular coordinates of a particle P by x, y, and z, we found
that the rectangular components of the velocity and acceleration of P equal,
respectively, the first and second derivatives with respect to t of the
corresponding coordinates. Thus,
.
.
.
vx 5 x
vy 5 y
vz 5 z
(11.28)
ay 5 ÿ
az 5 z̈
ax 5 ẍ
(11.29)
Component Motions
When the component ax of the acceleration depends only upon t, x, and/or
vx; when, similarly, ay depends only upon t, y, and/or vy; and az upon t, z,
and/or vz, Eq. (11.29) can be integrated independently. The analysis of the
given curvilinear motion then reduces to the analysis of three independent
rectilinear component motions [Sec. 11.4C]. This approach is particularly
effective in the study of the motion of projectiles [Sample Probs. 11.10
and 11.11].
y'
y
B
rB
Relative Motion of Two Particles
For two particles A and B moving in space (Fig. 11.28), we considered the relative motion of B with respect to A, or more precisely, with respect to a moving
frame attached to A and in translation with A [Sec. 11.4D]. Denoting the relative
position vector of B with respect to A by rB/A (Fig. 11.28), we have
rB 5 rA 1 rByA
(11.30)
O
rA
rB/A
A
x'
x
z'
z
Fig. 11.28
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Denoting the relative velocity and the relative acceleration of B with respect
to A by vB/A and aB/A, respectively, we also showed that
vB 5 vA 1 vB/A
(11.32)
aB 5 aA 1 aB/A
(11.33)
and
Tangential and Normal Components
It is sometimes convenient to resolve the velocity and acceleration of a particle
P into components other than the rectangular x, y, and z components. For a particle
P moving along a path contained in a plane, we attached to P unit vectors et
tangent to the path and en normal to the path and directed toward the center of
curvature of the path [Sec. 11.5A]. We then express the velocity and acceleration
of the particle in terms of tangential and normal components. We have
v 5 vet
(11.35)
and
y
C
a5
v2
a n = ρ en
at =
dv
et
dt
P
O
x
v2
dv
et 1 en
r
dt
(11.38)
where v is the speed of the particle and ρ is the radius of curvature of its path
[Sample Probs. 11.16, ,and 11.17]. We observed that, while the velocity v is
directed along the tangent to the path, the acceleration a consists of a component
at directed along the tangent to the path and a component an directed toward
the center of curvature of the path (Fig. 11.29).
Motion Along a Space Curve
For a particle P moving along a space curve, we defined the plane that most
closely fits the curve in the neighborhood of P as the osculating plane. This
plane contains the unit vectors et and en that define the tangent and principal
normal to the curve, respectively. The unit vector eb, which is perpendicular
to the osculating plane, defines the binormal.
Fig. 11.29
Radial and Transverse Components
eθ
er
r = re r
O
θ
Fig. 11.30
P
When the position of a particle P moving in a plane is defined by its polar
coordinates r and θ, it is convenient to use radial and transverse components
directed, respectively, along the position vector r of the particle and in the
direction obtained by rotating r through 90° counterclockwise [Sec. 11.5B].
We attached to P unit vectors er and eθ directed in the radial and transverse
directions, respectively (Fig. 11.30). We then expressed the velocity and acceleration of the particle in terms of radial and transverse components as
.
.
v 5 r e r 1 r θeθ
(11.42)
.2
$
..
$
(11.43)
a 5 (r 2 rθ )er 1 (rθ 1 2rθ)eθ
where dots are used to indicate differentiation with respect to time. The scalar
components of the velocity and acceleration in the radial and transverse
directions are therefore
.
.
vr 5 r
vθ 5 r θ
(11.44)
.
$
..
$
ar 5 r 2 rθ 2 aθ 5 rθ 1 2rθ
(11.45)
It is important to note that ar is not equal to the time derivative of vr and that aθ
is not equal to the time derivative of vθ [Sample Probs. 11.18, 11.19, and 11.20].
This chapter ended with a discussion of the use of cylindrical coordinates to define the position and motion of a particle in space.
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Review Problems
11.182 The motion of a particle is defined by the relation x 5 2t3 2 15t2 1
24t 1 4, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and
the total distance traveled when the acceleration is zero.
11.183 A drag car starts from rest and moves down the racetrack with
an acceleration defined by a 5 50 2 10t , where a and t are in
m/s2 and seconds, respectively. After reaching a speed of 125 m/s, a
parachute is deployed to help slow down the dragster. Knowing that
2
this deceleration is defined by the relationship a 5 20.02v , where v
is the velocity in m/s, determine (a) the total time from the beginning
of the race until the car slows back down to 10 m/s, (b) the total
distance the car travels during this time.
11.184 A particle moves in a straight line with the acceleration shown in the
figure. Knowing that the particle starts from the origin with
v0 5 22 m/s, (a) construct the v – t and x – t curves for 0 , t , 18 s,
(b) determine the position and the velocity of the particle and the
total distance traveled when t 5 18 s.
a (m /s 2)
6
2
8
12
– 0.75
t (s)
Fig. P11.184
11.185 The velocities of commuter trains A and B are as shown. Knowing that
the speed of each train is constant and that B reaches the crossing
10 min after A passed through the same crossing, determine (a) the
relative velocity of B with respect to A, (b) the distance between the
fronts of the engines 3 min after A passed through the crossing.
66 km/h
A
48 km/h
B
25°
Fig. P11.185
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11.186 Block B starts from rest and moves downward with a constant acceleration. Knowing that after slider block A has moved 400 mm is
velocity is 4 m/s, determine (a) the acceleration of A and B, (b) the
velocity and change in position of B after 2 s.
A
11.187 Collar A starts from rest at t 5 0 and moves downward with
a constant acceleration of 175 mm/s2. Collar B moves upward with a
constant acceleration, and its initial velocity is 200 mm/s. Knowing
that collar B moves through 500 mm between t 5 0 and t 5 2 s,
determine (a) the accelerations of collar B and block C, (b) the time
at which the velocity of block C is zero, (c) the distance through
which block C will have moved at that time.
B
11.188 A golfer hits a ball with an initial velocity of magnitude v0 at an
angle α with the horizontal. Knowing that the ball must clear the
tops of two trees and land as close as possible to the flag, determine
v0 and the distance d when the golfer uses (a) a six-iron with α 5 31°,
(b) a five-iron with α 5 27°.
Fig. P11.186
v0
a
B
70 m
d
10 m
Fig. P11.188
A
C
Fig. P11.187
B
A
30 m
14 m
12 m
11.189 As the truck shown begins to back up with a constant acceleration
of 1.2 m/s2, the outer section B of its boom starts to retract with a
constant acceleration of 0.48 m/s2 relative to the truck. Determine
(a) the acceleration of section B, (b) the velocity of section B when
t 5 2 s.
11.190 A velodrome is a specially designed track used in bicycle racing that
has constant radius curves at each end. Knowing that a rider starts
from rest at 5 (11.46 2 0.01878v2) m/s2, determine her acceleration
at point B.
50°
B
28 m
Fig. P11.189
.5
18
m
A
Fig. P11.190
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11.191 Sand is discharged at A from a conveyor belt and falls onto the top
of a stockpile at B. Knowing that the conveyor belt forms an angle
α 5 25° with the horizontal, determine (a) the speed v0 of the belt,
(b) the radius of curvature of the trajectory described by the sand
at point B.
11.192 The end point B of a boom is originally 5 m from fixed point A
when the driver starts to retract the boom with a constant radial
2
acceleration of
$ r̈ 5 21.0 m/s and lower it with a constant angular
acceleration θ 5 20.5 rad/s rad/s2. At t 5 2 s, determine (a) the
velocity of point B, (b) the acceleration of point B, (c) the radius of
curvature of the path.
v0
a
A
5.4 m
B
9m
Fig. P11.191
B
A
60°
Fig. P11.192
11.193 A telemetry system is used to quantify kinematic values of a ski
jumper immediately before she leaves the ramp. According to
.
2
the
. system r 5$ 150 m, r 5 2231.5 m/s, r̈ 5 23 m/s , θ 5 25°,
u 5 0.07 rad/s, u 5 0.06 rad/s . Determine (a) the velocity of the
skier immediately before she leaves the jump, (b) the acceleration of
the skier at this instant, (c) the distance of the jump d neglecting lift
and air resistance.
30°
3m
r
q
d
Fig. P11.193
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