# User manual | Free Fall ```Free Fall
Held down by shoulder
bars, you stop for a
moment at the top.
The supports under the
car are released, and
you’re in free fall,
dropping faster and
faster for a time interval
of 1.5 s. How far will
you drop and how fast
will you be going the
instant before the car
reaches the bottom?
➥ Look at the Example
Problem on page 105
CHAPTER
5
A Mathematical
Model of Motion
or pure heart-stopping excitement, nothing beats a ride on the
Demon Drop at Cedar Point Amusement Park. Enthusiasts
say its free-fall drop of 60 feet is the ultimate rush. Isn’t the
excitement of most of the rides the rapid and unexpected changes
in speed and direction? And what about those roller-coaster
lurches and sudden stops; the swaying and the hairpin turns?
Your favorite rides are probably the ones with the sharpest turns
and the most precipitous drops. Your ticket to ride entitles you to
What’s hard to imagine is that the Demon Drop and its neighboring roller coaster, the Millennium 2000, depend on the same
basic principles of physics. Once your Demon Drop car reaches
the top of the tower, no motors, engines, pulleys, or other energy
source interferes with your fall. The same is true for roller
coasters. Initially, cables pull the coaster cars to the top of the first
hill. From there the laws of physics take charge, propelling the
rides downhill, up again, through loops and spirals at speeds of
37 km/h (60 mph) or more. In fact, many amusement park rides,
from the gently circling carousel and Ferris wheel to the tilt-owhirl and roller coaster, offer rides that can be specified and
described in mathematical terms.
The vectors you learned about in Chapter 4, together with the
concepts of velocity and acceleration, can help explain why amusement parks are fun. They also explain how such hair-raising rides
as the Demon Drop and Millennium 2000 roller coasters simulate danger while remaining safe.
F
WHAT YOU’LL LEARN
•
•
You will continue your
study of average and
instantaneous velocity,
and acceleration.
You will use graphs and
equations to solve problems
involving moving objects,
including freely falling
objects.
WHY IT’S IMPORTANT
•
The rapid pace of life today
means that modern cars,
planes, elevators, and other
people-moving vehicles
often are designed to accelerate quickly from rest to the
highest speed deemed safe,
and then continue at that
speed until they reach their
destinations.
PHYSICS
To find out more about motion,
visit the Glencoe Science Web site
at science.glencoe.com
81
5.1
OBJ ECTIVES
• Interpret graphs of position
versus time for a moving
object to determine the
velocity of the object.
•
Describe in words the
information presented in
graphs and draw graphs
from descriptions of motion.
•
Write equations that
describe the position of an
object moving at constant
velocity.
TABLE 5–1
Position versus Time
Time,
t (s)
Position,
x (m)
0.0
10
1.0
12
2.0
18
3.0
26
4.0
36
5.0
43
6.0
48
Graphing Motion in
One Dimension
Y
ou have learned how to describe motion in terms of
words, sketches, and motion diagrams. In this chapter,
you’ll learn to represent one-dimensional motion by means of a graph of
position versus time. Such a graph presents information not only about the
displacement of an object, but also about its velocity.
The other tool that is useful for certain kinds of motion is an equation
that describes an object’s displacement versus time. You can use a
position-time graph and this equation to analyze the motion of an object
mathematically and to make predictions about its position, velocity, and
acceleration.
Position-Time Graphs
How could you make a graph of the position of the Demon Drop
at various times? Such a graph would be a position-time graph. You
will learn to make a p-t graph for the Demon Drop, but first, consider a
simpler example. A physics student uses a camcorder to record the
motion of a running back as he runs straight down the football field to
make a touchdown. She records one frame each second and produces
the motion diagram shown in Figure 5–1. From the motion diagram,
the physics student obtains the data in Table 5–1. Because she chooses
the x-coordinate axis, the symbol x in this problem represents distance
from his own goal line.
Notice the origin of the coordinate system in Figure 5–1. Time was
set to zero when the running back began to move with the ball, but the
origin of the x-axis was not chosen to be his initial position. Instead, he
began 10 m from the origin.
Two graphs of the running back‘s motion are shown in Figure 5–2. In
the first graph, only the recorded positions are shown. In the second
graph, a curve connects each of the recorded points. These lines represent
our best guess as to where the running back was in between the recorded
points. You can see that this graph is not a picture of the path taken by the
ball carrier as he was running; the graph is curved, but the path that he
took down the field was not.
Begin
FIGURE 5–1 The football player
began at 10 m and ran in the
positive x direction.
82
End
+x
10 m
A Mathematical Model of Motion
20 m
30 m
40 m
50 m
x (m)
40
40
30
30
20
20
10
0
FIGURE 5–2 Only the plotted
points are known. The lines joining the points are best guesses
as to the position of the running
back.
x (m)
1
2
3
4
5
a
t (s)
6
10
0
1
2
3
4
5
t (s)
6
b
What is an instant of time? How long did the running back spend
at any location? Each position has been linked to a time, but how long
did that time last? You could say “an instant,” but how long is that? If an
instant lasts for any finite amount of time, then, because the running back
would be at the same position during that time, he would be at rest. But
a moving object cannot be at rest; thus, an instant is not a finite period of
time. This means that an instant of time lasts zero seconds. The symbol x
represents the instantaneous position of the running back at a particular
instant of time.
Using a Graph to Find Out Where and When
When did the running back reach the 30-m mark? Where was he
4.5 s after he started running? These questions can be answered easily
with a position-time graph, as you will see in the following example
problem. Note that the questions are first restated in the language of
physics in terms of positions and times.
Example Problem
Data from a Position-Time Graph
When did the running back reach the 30-m mark? Where was he
after 4.5 seconds?
Strategy:
Restate the questions:
x (m)
Question 1: At what time was the position of the object equal
to 30 m?
40
Question 2: What was the position of the object at 4.5 s?
To answer question 1, examine the graph to find the intersection
of the curve with a horizontal line at the 30-m mark.
To answer question 2, find the intersection of the curve with a
vertical line at 4.5 s (halfway between 4 s and 5 s on this graph).
30
20
10
t (s)
0 1
2
3
4 5 6
The two intersections are shown on the graph.
5.1 Graphing Motion in One Dimension
83
Graphing the Motion of Two or More Objects
Pictorial and graphical representations of the running back and two
players, A and B, on the opposing team are shown in Figure 5–3. When
and where would each of them have a chance to tackle the running
back? First, you need to restate this question in physics terms: At what
time do two objects have the same position? On a position-time graph,
when do the curves representing the two objects intersect?
FIGURE 5–3 When and where
can A and B tackle the running
back?
x (m)
Running
back
B
A
B
40
30
A
+x
20
t (s)
10
0
1
2
3
4
5
6
Running back and A meet
+x
Running back and B meet
+x
Example Problem
Interpreting Position-Time Graphs
Where and when do defenders A and B have a chance to tackle
the running back?
Strategy:
In Figure 5–3, the intersections of the curves representing the
motion of defenders A and B with the curve representing the
motion of the running back are points.
Defender A intersects with running back in 4 s at about 35 m.
Defender B intersects in 5 s at about 42 m.
From Graphs to Words and Back Again
To interpret a position-time graph in words, start by finding the position of the object at t 0. You have already seen that the position of the
object is not always zero when t 0. Then, examine the curve to see
whether the position increases, remains the same, or decreases with
increasing time. Motion away from the origin in a positive direction has
a positive velocity, and motion in the negative direction has a negative
velocity. If there is no change in position, then the velocity is zero.
84
A Mathematical Model of Motion
Example Problem
Describing Motion from a Position-Time Graph
Describe the motion of the players in Figure 5–3.
Strategy:
The running back started at the 10-m mark and moved in the
positive direction, that is, with a positive velocity.
Defender A started at 25 m. After waiting about 1.5 s, he also
moved with positive velocity.
Defender B started at 45 m. After 3 s, he started running in the
opposite direction, that is, with a negative velocity.
Practice Problems
1. Describe in words the motion of the four walkers shown by the
four lines in Figure 5–4. Assume the positive direction is east
and the origin is the corner of High Street.
2. Describe the motion of the car shown in Figure 5–5.
graphed in Figure 5–5.
a. When was the car 20 m west of the origin?
b. Where was the car at 50 s?
c. The car suddenly reversed direction. When and where did
that occur?
d (m)
East
x
A
D
30
20
10
B
t
High St.
C
FIGURE 5–4
B
–20
–30
E
A
–10
D
West
0
C
0
10
20
30
40
50
60
70
t (s)
FIGURE 5–5
Uniform Motion
If an airplane travels 75 m in a straight line in the first second of its
flight, 75 m in the next second, and continues in this way, then it is
moving with uniform motion. Uniform motion means that equal displacements occur during successive equal time intervals. A motion diagram and a position-time graph can be used to describe the uniform
motion of the plane.
5.1 Graphing Motion in One Dimension
85
280
The line drawn through the points representing the position of the
plane each second is a straight line. Recall from Chapter 2 that one of
the properties of a straight line is its slope. To find the slope, take the
ratio of the vertical difference between two points on the line, the rise,
to the horizontal difference between the same points, the run.
rise
slope run
d (m)
240
200
d1
160
120
d0
80
40
0
0
t0
t1
1
2
d
t (s)
3
+x
FIGURE 5–6 The regular
change in position shows that
this airplane is moving with
uniform velocity.
FIGURE 5–7 You can tell in the
graph below that the riders are
traveling at different velocities
because the three lines representing their velocities have
different slopes.
Figure 5–6 shows how to determine the slope using the points at 1 s
and 2 s. Any set of points would produce the same result because a
straight line has a constant slope.
When the line is a position-time graph, then rise d and run t,
so the slope is the average velocity.
d1 d0
d
slope v t
t1 t0
Thus, on a position-time graph, the slope of a straight line passing
through the points on the graph at times t0 and t1 is the average velocity between any two times. In Figure 5–6, notice that when t0 is 1 s and
t1 is 2 s, d0 equals 115 m and d1 equals 190 m. The average velocity of
the airplane v (190 m 115 m)/(2 s 1 s) 75 m/s.
Note that the average velocity is not d/t, as you can see by calculating
that ratio from the coordinates of the points on the graph. Check this
for yourself using Figure 5–6. You will find that when t 2 s and
d 190 m, the ratio d/t is 95 m/s, not 75 m/s, as calculated in the
previous equation.
What do position-time graphs look like for different velocities? Look
at the graph of three bike riders in Figure 5–7. Note the initial position
of the riders. Rider A has a displacement of 4.0 m in 0.4 s, so the average velocity of rider A is the following.
d
4.0 m
2.0 m (2.0 m)
vA 10 m/s
0.4 s 0 s
t
0.4 s
What is the average velocity, vB, of rider B? The slope of the line is less
than that for rider A, so the magnitude of the velocity of rider B should
be smaller. With a displacement of 4.0 m in 0.6 s, the speed is 6.7 m/s,
less than that of A, as expected.
d (m)
6
A
4
B
2
t (s)
0
–2
–4
C
–6
86
A Mathematical Model of Motion
0.2
0.4
0.6
0.8
What about rider C? Although rider C moves 4.0 m in 0.6 s, her displacement, which is the final position minus the initial position, is
4.0 m. Thus, her average velocity is vC 6.7 m/s. This tells you that
rider C is moving in the negative direction. A line that slants downward
and has a negative slope represents a negative average velocity. Recall
that a velocity is negative if the direction of motion is opposite the direction you chose to be positive.
Practice Problems
4. For each of the position-time graphs shown in Figure 5–8,
a. write a description of the motion.
b. draw a motion diagram.
c. rank the average velocities from largest to smallest.
5. Draw a position-time graph for a person who starts on the positive side of the origin and walks with uniform motion toward
the origin. Repeat for a person who starts on the negative side of
the origin and walks toward the origin.
6. Chris claims that as long as average velocity is constant, you can
write v d/t. Use data from the graph of the airplane’s motion
in Figure 5–6 to convince Chris that this is not true.
7. Use the factor-label method to convert the units of the following
average velocities.
a. speed of a sprinter: 10.0 m/s into mph and km/h
b. speed of a car: 65 mph into km/h and m/s
c. speed of a walker: 4 mph into km/h and m/s
8. Draw a position-time graph of a person who walks one block at
a moderate speed, waits a short time for a traffic light, walks the
next block slowly, and then walks the final block quickly. All
blocks are of equal length.
Using an Equation to Find Out
Where and When
Uniform motion can be represented by an algebraic equation. Recall
from Chapter 3 that the average velocity was defined in this way.
d
t
d d
t1 t0
1
0
Average Velocity v Note the absence of bold-face type. All algebra is done using the components of vectors and not the vectors themselves. Assume that you’ve
chosen the origin of the time axis to be zero, so that t0 0. Then, t0 can
be eliminated and the equation rearranged.
d
C
B
A
D
t
FIGURE 5–8
Pocket Lab
Uniform or Not?
Set up a U-channel on a book
so that it acts as an inclined
ramp, or make a channel from
two metersticks taped together
along the edges. Release a
steel ball so that it rolls down
the ramp. Using a stopwatch,
measure the time it takes the
ball to roll 0.40 m.
Analyze and Conclude Write
a brief description of the
motion of the ball. Predict how
much time it would take the
ball to roll 0.80 m. Explain
d1 d0 vt1
5.1 Graphing Motion in One Dimension
87
F.Y.I.
A stroboscope provides
intermittent illumination
of an object so that the
object’s motion, rotary
speed, or frequency of
vibration may be studied.
The stroboscope causes an
object to appear to slow
down or stop by producing
illumination in short bursts
duration at a frequency
selected by the user.
The equation can be made more general by letting t be any value of
t1 and d be the value of the position at that time. In addition, the
distinction between velocity and average velocity is not needed because
you will be working with the special case of constant velocity. This means
that the average velocity between any two times will be the same as the
constant instantaneous velocity, v v. The symbol v represents velocity,
and d0 represents the position at t 0. The following equation is then
obtained for the position of an object moving at constant velocity.
Position with Constant Velocity d d0 vt
The equation involves four quantities: the initial position, d0, the
constant velocity, v, the time, t, and the position at that time, d. If you
are given three of these quantities, you can use the equation to find the
fourth. When a problem is stated in words, you have to read it carefully
to find out which three are given and which is unknown. When problems are given in graphical form, the slope of the curve tells you the
constant velocity, and the point where the curve crosses the t 0 line is
the initial position. The following example problem illustrates the use
of both a graph and the equation in solving a problem.
Example Problem
Finding Position from a Graph and an Equation
Write the equation that describes the motion of the airplane graphed
in Figure 5–6, and find the position of the airplane at 2.5 s.
Known:
Unknown:
v 75 m/s
d?
d0 40 m
t 2.5 s
Strategy:
Calculations:
The constant velocity is 75 m/s.
d d0 vt
The curve intersects the t 0 line at 40 m,
d 40 m (75 m/s)(2.5 s)
so the initial position is 40 m.
230 m
You know the time, so the equation to use is
d d0 vt.
• Is the unit correct? m/s s results in m.
• Does the sign make sense? It is positive, as it should be.
• Is the magnitude realistic? The result agrees with the value shown
on the graph.
88
A Mathematical Model of Motion
Practice Problems
9. Consider the motion of bike rider A in Figure 5–7.
a. Write the equation that represents her motion.
b. Where will rider A be at 1.0 s?
10. Consider the motion of bike rider C in Figure 5–7.
a. Write the equation that represents her motion.
b. When will rider C be at 10.0 m?
11. A car starts 2.0 102 m west of the town square and moves
with a constant velocity of 15 m/s toward the east. Choose a
coordinate system in which the x-axis points east and the origin
is at the town square.
a. Write the equation that represents the motion of the car.
b. Where will the car be 10.0 min later?
c. When will the car reach the town square?
12. At the same time the car in problem 11 left, a truck was 4.0 102 m
east of the town square moving west at a constant velocity of
12 m/s. Use the same coordinate system as you did for problem 11.
a. Draw a graph showing the motion of both the car and the truck.
b. Find the time and place where the car passed the truck using
both the graph and your two equations.
EARTH SCIENCE
CONNECTION
Space Probes Scientists
use other planets’ gravity
to alter the path of a
space probe and increase
its speed. The probe is
programmed to pass near
a planet. The planet’s
gravity bends the probe’s
trajectory and propels it
away. In 1974, Mariner 10
swung by Venus to gain
energy for three passes by
Mercury. In 1992, Jupiter’s
gravity allowed Ulysses to
make a hairpin turn on its
way to take photos of the
sun’s south pole.
5.1 Section Review
1. You drive at constant speed toward
the grocery store, but halfway there
you realize you forgot your list. You
quickly turn around and return home
at the same speed. Describe in words
the position-time graph that would
2. A car drives 3.0 km at a constant
speed of 45 km/h. You use a coordinate system with its origin at the point
where the car started and the direction
of the car as the positive direction.
Your friend uses a coordinate system
with its origin at the point where the
car stopped and the opposite direction as the positive direction. Would
the two of you agree on the car’s
position? Displacement? Distance?
Velocity? Speed?
3. Write equations for the motion of the
car just described in both coordinate
systems.
4.
Critical Thinking A police officer
clocked a driver going 20 mph over
the speed limit just as the driver
passed a slower car. He arrested both
drivers. The judge agreed that both
were guilty, saying, “If the two cars
were next to each other, they must
have been going the same speed.”
Are the judge and police officer correct? Explain with a sketch, a motion
diagram, and a position-time graph.
5.1 Graphing Motion in One Dimension
89
5.2
Graphing Velocity in
One Dimension
Y
OBJ ECTIVES
• Determine, from a graph
of velocity versus time, the
velocity of an object at a
specified time.
•
•
Interpret a v-t graph to find
the time at which an object
has a specific velocity.
Calculate the displacement
of an object from the area
under a v-t curve.
FIGURE 5–9 The displacement
during equal time intervals
increases as velocity increases.
The graph in b shows that the
slope of the tangent to the curve
at any point is the instantaneous
velocity at that time.
90
ou’ve learned how to draw a position-time graph
for an object moving at a constant velocity, and how
to use that graph to write an equation to determine the
velocity of the object. You also have learned how to use both the graph and
the equation to find an object’s position at a specified time, as well as the
time it takes the object to reach a specific position. Now you will explore
the relationship between velocity and time when velocity is not constant.
Determining Instantaneous Velocity
What does a position-time graph look like when an object is going
faster and faster? Figure 5–9a shows a different position-time record of
an airplane flight. The displacements for equal time intervals in both
the motion diagram and the graph get larger and larger. This means that
the average velocity during each time interval, v d/t, also gets
larger and larger. The motion is certainly not uniform. The instantaneous velocity cannot equal the average velocity.
How fast was the plane going at 1.5 s? The average velocity is the
slope of the straight line connecting any two points. Using Figure 5–9a,
you can see that the average velocity between 1 s and 2 s is d/t, that
is (10 m 4 m)/(1 s), or 6 m/s. But the velocity could have changed
within that second. To be precise, more data are needed. Figure 5–9b
shows the slope of the line connecting 1.25 s and 1.75 s. Because this
time interval is half of the 1-s time interval used previously, the average
velocity is probably closer to the instantaneous velocity at 1.5 s. You
could continue the process of reducing the time interval and finding the
ratio of the displacement to the time interval. Each time you reduce the
time interval, the ratio is likely to be closer to the quantity called the
d (m)
d (m)
20
18
16
14
12
10
8
6
4
2
0
20
18
16
14
12
10
8
6
4
2
0
d
d
d
t (s)
0
a
A Mathematical Model of Motion
1
2
0
3
+x
b
t (s)
1
2
3
Slope of this line is the
instantaneous velocity
instantaneous velocity. Finally, you would find that the slope of the line
tangent to the position-time curve at a specific time is the instantaneous
velocity, v, at that time.
Pocket Lab
Velocity-Time Graphs
Now that you know what the velocity at an instant of time is and how
to find it, you will be able to draw a graph of the velocity versus time of
the airplane whose motion is not constant. Just as average velocity
arrows on a motion diagram are drawn in red, the curves on a velocitytime (v-t) graph are also shown in red.
Motion diagrams and a velocity-time graph for the two airplanes,
one with constant velocity and the other with increasing velocity, are
shown in Figure 5–10 for a 3-s time interval during their flights. Note
that the velocities are given in m/s. The velocity of one airplane is
increasing, while the velocity of the other airplane is constant. Which
line on the graph represents the plane with uniform motion? Uniform
motion, or constant velocity, is represented by a horizontal line on a
v-t graph. Airplane B is traveling with a constant velocity of 75 m/s, or
270 km/h. The second line increases from 70 m/s to 82 m/s over a
3-s time interval. At 1.5 s, the velocity is 76 m/s. This is the slope of
the tangent to the curve on a position-time graph at 1.5 s for the flight
of airplane A.
What would the v-t graph look like if a plane were going at constant
speed in the opposite direction? As long as you don’t change the direction of the coordinate axis, the velocity would be negative, so the graph
would be a horizontal line below the t-axis.
What can you learn from the intersection of the two lines on the
graph? When the two v-t lines cross, the two airplanes have the same
velocity. The planes do not necessarily have the same position, so they
do not meet at this time. Velocity-time graphs give no information
about position, although, as you will learn in the next section, you can
use them to find displacement.
A Ball Race
Assemble an inclined ramp
from a piece of U-channel or
two metersticks taped together.
Make a mark at 40 cm from
the top and another at 80 cm
from the top. If two balls are
released at the same instant,
one ball from the top and the
other ball at 40 cm, will the
balls get closer or farther apart
as they roll down the ramp?
Why? Try it. Now, release one
ball from the top and then
release another ball from the
top as soon as the first ball
reaches the 40-cm mark.
Analyze and Conclude
terms of velocities. Do the balls
ever have the same velocities
as they roll down the hill? Do
they have the same
acceleration?
Math Handbook
T
o review calculating the
area under a graph, see the
Math Handbook, Appendix A,
page 743.
v (m/s)
82
80
VB
78
A
76
VA
B
74
72
t (s)
70
0
1
2
+x
FIGURE 5–10 The lines on the
graph and the motion diagrams
are two different ways of representing both constant and
increasing velocity.
3
5.2 Graphing Velocity in One Dimension
91
v (m/s)
82
Displacement from a Velocity-Time Graph
For an object moving at constant velocity,
d
v v , so d vt.
t
80
78
76
74
72
t (s)
70
0
1
2
3
FIGURE 5–11 The displacement
during a given time interval is
the area under the curve of a
v-t graph.
As you can see in Figure 5–11, v is the height of the curve above the
t-axis, while t is the width of the shaded rectangle. The area of the rectangle, then, is vt, or d. You can find the displacement of the object by
determining the area under the v-t curve.
If the velocity is constant, the displacement is proportional to the
width of the rectangle. Thus, if you plot the displacement versus time,
you will get a straight line with slope equal to velocity. If the velocity is
increasing, then the area of the rectangle increases in time, so the slope
of a displacement-versus-time graph also increases.
Example Problem
Finding the Displacement of an Airplane
from Its v-t Graph
Find the displacement of the plane in Figure 5–11 that is moving at
constant velocity after
a. 1.0 s.
b. 2.0 s.
c. 3.0 s.
Compare your results to the original position-time graph in Figure 5–6.
Known:
Unknown:
v 75 m/s
t 1.0 s, 2.0 s, and 3.0 s
d at 1.0 s, 2.0 s, and 3.0 s
Strategy:
Calculations:
The displacement is the area under
the curve, or d vt.
a. d vt (75 m/s)(1.0 s) 75 m
b. d vt (75 m/s)(2.0 s) 150 m
c. d vt (75 m/s)(3.0 s) 225 m 230 m
• Are the units correct? m/s s m.
• Do the signs make sense? They are all positive, as they should be.
• Are the magnitudes realistic? The calculated positions are different from those on the position-time graph: 115 m, 190 m, and
265 m. The differences occur because the initial position of the
plane at t 0 is 40 m, not zero. You must add the displacement
of the airplane at t 0 to the value calculated at each time.
92
A Mathematical Model of Motion
Practice Problems
13. Use Figure 5–10 to determine the velocity of the airplane that
is speeding up at
a. 1.0 s.
b. 2.0 s.
c. 2.5 s.
14. Use the factor-label method to convert the speed of the airplane
whose motion is graphed in Figure 5–6 (75 m/s) to km/h.
15. Sketch the velocity-time graphs for the three bike riders in
Figure 5–7.
16. A car is driven at a constant velocity of 25 m/s for 10.0 min. The
car runs out of gas, so the driver walks in the same direction at
1.5 m/s for 20.0 min to the nearest gas station. After spending
10.0 min filling a gasoline can, the driver walks back to the car
at a slower speed of 1.2 m/s. The car is then driven home at 25
m/s (in the direction opposite that of the original trip).
a. Draw a velocity-time graph for the driver, using seconds as
your time unit. You will have to calculate the distance the
driver walked to the gas station in order to find the time it
took the driver to walk back to the car.
b. Draw a position-time graph for the problem using the areas
under the curve of the velocity-time graph.
Pocket Lab
Bowling Ball
Displacement
Take a bowling ball and three
stopwatches into the hallway.
Divide into three groups. Have
all timers start their watches
when the ball is rolled. Group 1
should stop its watch when the
ball has gone 10 m, group 2
should stop its watch when the
ball has rolled 20 m, and group
3 should stop its watch when
the ball has rolled 30 m.
Analyze and Conclude
Record the data and calculate
the average speed for each
distance. Could the average
speed for 30 m be used to
predict the time needed to roll
100 m? Why or why not?
5.2 Section Review
at which an object had a specified
velocity.
1. What information can you obtain
from a velocity-time graph?
2. Two joggers run at a constant velocity
of 7.5 m/s toward the east. At time
t = 0, one is 15 m east of the origin;
the other is 15 m west.
a. What would be the difference(s)
in the position-time graphs of
their motion?
b. What would be the difference(s)
in their velocity-time graphs?
3. Explain how you would use a
velocity-time graph to find the time
4. Sketch a velocity-time graph for a car
that goes 25 m/s toward the east for
100 s, then 25 m/s toward the west
for another 100 s.
5.
Critical Thinking If the constant
velocity on a v-t graph is negative,
what is the sign of the area under
the curve? Explain in terms of the
displacement of the object whose
motion is represented on the graph.
5.2 Graphing Velocity in One Dimension
93
5.3
Acceleration
I
OBJ ECTIVES
• Determine from the curves
on a velocity-time graph
both the constant and
instantaneous acceleration.
•
Determine the sign of
acceleration using a v-t
graph and a motion
diagram.
•
n Chapter 3, you learned how to use a motion diagram to get a feel for the average acceleration of an
object. This method is illustrated in Figure 5–12b for
the motion of two airplanes. Airplane A travels with non-uniform velocity,
so the change in velocity, and thus the acceleration, is in the same direction as the velocity. Both velocity and acceleration have a positive sign.
Airplane B travels with uniform velocity, so its acceleration is zero.
Determining Average Acceleration
Average acceleration is the rate of change of velocity between t0 and t1,
and is represented by the following equation.
Calculate the velocity and
the displacement of an
object undergoing constant
acceleration.
v
t
v v
t1 t0
1
0
Average Acceleration a You can find this ratio by determining the slope of the velocity-time
graph in the same way you found velocity from the slope of the
position-time graph. For example, Figure 5–12a shows that in a 1-s time
interval, the velocity of plane A increases by 4 m/s. That is, v 4 m/s
and t 1 s, so a v/t 4 m/s2.
Constant and Instantaneous Acceleration
FIGURE 5–12 Graphs and
motion diagrams are useful in
differentiating motion having
uniform velocity and motion that
is accelerated.
v (m/s)
82
80
Recall that an object undergoes uniform motion, or constant velocity, if the slope of the position-time graph is constant. Does the slope
of the velocity-time graph for the accelerating airplane of Figure 5–12a
change? No, it rises by 4 m/s every second. This type of motion, which
can be described by a constant slope on a velocity-time graph, is called
constant acceleration.
What if the slope of a v-t graph isn’t constant? You learned that you
could find the instantaneous velocity by finding the slope of the tangent
to the curve on the position-time graph. In the same way, you can find
instantaneous acceleration, a, as the slope of the tangent to the curve
on a velocity-time graph. Instantaneous acceleration is the acceleration
of an object at an instant of time. Consider the velocity-time graph in
the example problem.
78
A
76
Airplane A:
B
74
Begin
End
a>0
+x
72
70
a
94
0
1
2
3
t (s)
Airplane B:
b
A Mathematical Model of Motion
End
Begin
a=0
Example Problem
How would you describe the sprinter’s velocity and
acceleration as shown on the graph?
12
v (m/s)
Determining Velocity and Acceleration
from a Graph
Strategy:
From the graph, note that the sprinter’s
velocity starts at zero, increases rapidly
for the first few seconds, and then, after
reaching about 10 m/s, remains almost
constant.
Draw a tangent to the curve at two different
times, t 1 s and t 5 s.
The slope of the lines at 1 s and 5 s is the
acceleration at those times.
6
0
0
5
t (s)
10
Calculations:
12.0 m/s 3.0 m/s
rise
At 1 s, a ;
2.5 s 0.0 s
run
a 3.6 m/s2
10.7 m/s 10.0 m/s
At 5 s, a 10 s 0 s
a 0.07 m/s2
The acceleration is 3.6 m/s2 at 1 s, and
0.07 m/s2 at 5 s. It is larger before 1 s and
smaller after 5 s. The acceleration is not
constant.
The Zero Gravity Trainer
The zero-gravity trainer is an aircraft at
NASA’s Johnson Space Center designed for
use in simulating zero-gravity conditions such
as those experienced aboard spacecraft orbiting Earth. The plane is a modified Boeing 707
that mimics the free-fall environment aboard
a spacecraft as it flies a series of parabolashaped courses that take the crew from
12 000 m and back down again—all in less
than two minutes! These short spurts of simulated weightlessness enable the astronauts to
practice eating, drinking, and performing a
variety of tasks that they will carry out during
future missions. Because of the rapid ascents
and descents of the aptly nicknamed Vomit
Comet, training sessions are generally limited
to one or two hours.
During a run, the four-engine turbo jet
accelerates from 350 knots indicated airspeed
(KIA) to about 150 KIA at the top of the
parabola. There, the pilot adjusts the jet’s
engines so that speed is constant. Then, the
jet pitches over until the plane is descending
again at 350 KIA. During the ascent and
descent, acceleration is approximately 1.8 g.
Thinking Critically How are the Demon Drop
amusement park ride and a ride in the zerogravity trainer alike?
5.3 Acceleration
95
Positive and Negative Acceleration
You’ve considered the motion of an accelerating airplane and a
sprinter and found that, in both cases, the object’s velocity was positive
and increasing, and the sign of the acceleration was positive. Now consider a ball being rolled up a slanted driveway. What happens? It slows
down, stops briefly, then rolls back down the hill at an increasing speed.
Examine the two graphs in Figure 5–13 that represent the ball’s motion
and interpret them in the following example problem.
v (m/s)
3
v (m/s)
3
Case A
0
0
–3
0
t (s)
10
5
–3
+x
FIGURE 5–13 The sign of the
acceleration depends upon the
chosen coordinate system.
Case B
Begin
End
Same
point
0
+x
5
Begin
End
a<0
t (s)
10
Same
point
a>0
Example Problem
Finding the Sign of Acceleration
Describe the motion of the ball shown in Figure 5–13. What is
the difference between the two cases? What is the sign of the ball’s
acceleration? What is the magnitude of the ball’s acceleration?
Strategy:
In each case, the coordinate axis is parallel to the surface. In case A,
the ball initially moved in the direction of the positive axis; thus,
the sign of the velocity is positive. In case B, the axis was chosen in
the opposite direction; thus, the sign of the velocity is negative.
In each case, the ball started with a speed of 2.5 m/s. In both
cases, the ball slows down, reaches zero velocity, then speeds up
in the opposite direction.
To find the ball’s acceleration, use the motion diagrams. Subtract an
earlier velocity vector from a later one. Whether the ball is moving
uphill or downhill, the acceleration vector points to the left. In case
A, the positive axis points to the right, but the acceleration vector
points in the opposite direction. Therefore, the acceleration is negative. In case B, the positive axis points to the left, and the acceleration points in the same direction. The acceleration is positive.
Find the magnitude of the acceleration from the slopes of
the graphs.
96
A Mathematical Model of Motion
Calculations:
For case A, the ball slows down in the first 5 s.
v v1 v0 0.0 m/s 2.5 m/s 2.5 m/s
a v/t
a (2.5 m/s)/(5.0 s) 0.50 m/s2
During the next 5 s, the ball speeds up in the negative direction.
v 2.5 m/s 0.0 m/s 2.5 m/s
Again, a 0.50 m/s2
Check case B for yourself. Because the axis for case B was chosen in the
opposite direction, you will find that a is 0.50 m/s2, both when the
ball is slowing down and when it is speeding up.
Practice Problems
17. An Indy 500 race car’s velocity increases from 4.0 m/s to
36 m/s over a 4.0-s time interval. What is its average
acceleration?
18. The race car in problem 17 slows from 36 m/s to 15 m/s
over 3.0 s. What is its average acceleration?
19. A car is coasting backwards downhill at a speed of 3.0 m/s
when the driver gets the engine started. After 2.5 s, the car is
moving uphill at 4.5 m/s. Assuming that uphill is the positive
direction, what is the car’s average acceleration?
20. A bus is moving at 25 m/s when the driver steps on the brakes
and brings the bus to a stop in 3.0 s.
a. What is the average acceleration of the bus while braking?
b. If the bus took twice as long to stop, how would the acceleration compare with what you found in part a?
21. Look at the v-t graph of the toy train in Figure 5–14.
a. During which time interval or intervals is the speed constant?
b. During which interval or intervals is the train’s acceleration
positive?
c. During which time interval is its acceleration most negative?
22. Using Figure 5–14, find the average acceleration during the
following time intervals.
a. 0 to 5 s
b. 15 to 20 s
c. 0 to 40 s
v (m/s)
12
10
8
6
4
2
0
0
10
20
30
t (s)
40
FIGURE 5–14
Acceleration when instantaneous velocity is zero What happens to the acceleration when v = 0, that is, when the ball in the example problem stops and reverses direction? Consider the motion diagram
that starts when the ball is still moving uphill and ends when the ball is
5.3 Acceleration
97
Uphill
v
Begin
a
At top
v=0
a
Downhill
v
moving back downhill as in Figure 5–15. The acceleration is downhill
both before and after the ball stops. What happens at the instant when
the ball’s instantaneous velocity is zero? Remember that the velocity is
zero only at an instant of time, not for an instant. That is, the time interval over which the velocity is zero is itself zero. Thus, the acceleration
points downhill as the ball reaches the top of the hill, and it continues
to point downhill as the ball moves back downhill.
End
a
FIGURE 5–15 The acceleration
of the ball is always downward.
HELP WANTED
AIR-TRAFFIC
CONTROLLER
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communicate clearly and
decisively while juggling
dozens of details in your
mind, you might be the
person we are seeking.
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of college and the ability
to pass the Federal Civil
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is based on job performance and completion
of on-the-job training
programs. Knowledge of
foreign languages is a
plus. For information,
Information Center of
The U.S. Office of
Personnel Management.
98
Calculating Velocity from Acceleration
You learned that you could use the definition of velocity to find the
position of an object moving at constant velocity. In the same way, you
can find the velocity of the object by rearranging the definition of average acceleration.
v
v1 v0
a t
t1 t0
Assume that t0 0. At that time, the object has a velocity v0. Let t be
any value of t1 and v be the value of the velocity at that time. Because
only one-dimensional, straight-line motion with constant acceleration
will be considered, a can be used instead of a. After making these substitutions and rearranging the equation, the following equation is
obtained for the velocity of an object moving at constant acceleration.
Velocity with Constant Acceleration
v v0 at
You also can use this equation to find the time at which a constantly
accelerating object has a given velocity, or, if you are given both a velocity
and the time at which it occurred, you can calculate the initial velocity.
Practice Problems
23. A golf ball rolls up a hill toward a miniature-golf hole. Assign
the direction toward the hole as being positive.
a. If the ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2, what is its velocity after 2.0 s?
b. If the constant acceleration continues for 6.0 s, what will be
its velocity then?
c. Describe in words and in a motion diagram the motion of
the golf ball.
24. A bus, traveling at 30.0 km/h, speeds up at a constant rate of
3.5 m/s2. What velocity does it reach 6.8 s later?
25. If a car accelerates from rest at a constant 5.5 m/s2, how long
will it need to reach a velocity of 28 m/s?
26. A car slows from 22 m/s to 3.0 m/s at a constant rate of
2.1 m/s2. How many seconds are required before the car
is traveling at 3.0 m/s?
A Mathematical Model of Motion
v
Displacement Under Constant Acceleration
You know how to use the area under the curve of a velocity-time
graph to find the displacement when the velocity is constant. The same
method can be used to find the displacement when the acceleration is
constant. Figure 5–16 is a graph of the motion of an object accelerating constantly from v0 to v. If the velocity had been a constant, v0, the
displacement would have been v0t, the area of the lightly shaded rectangle. Instead, the velocity increased from v0 to v. Thus, the displacement is increased by the area of the triangle, 1/2(v – v0)t. The total displacement, then, is the sum of the two.
d v0t 1/2(v – v0)t
When the terms are combined, the following equation results.
d 1/2(v v0)t
If the initial position, d0, is not zero, then this term must be added to
give the general equation for the final position.
v
v0
t
0
0
FIGURE 5–16 The area under
the curve of a velocity versus time
graph equals the displacement.
Final Position with Constant Acceleration d d0 1/2(v v0)t
Frequently, the velocity at time t is not known, but because v v0 at,
you can substitute v0 at for v in the previous equation and obtain the
following equation.
d d0 1/2(v0 v0 at)t
When the terms are combined, the following equation results.
Final Position with Constant Acceleration d d0 v0t 1/2at2
Note that the third equation involves position, velocity, and time, but
not acceleration. The fifth equation involves position, acceleration, and
time, but not velocity. Is there an equation that relates position, velocity,
and acceleration, but doesn’t include time? To find that equation, start
with the following equations.
d d0 1/2(v v0)t and v v0 at
Solve the second equation for t.
t (v – v0)/a
Substitute this into the equation for displacement.
d d0 1/2(v0 v)(v – v0)/a
This equation can be solved for the final velocity.
Final Velocity with Constant Acceleration v2 v02 2a(d – d0)
The four equations that have been derived for motion under constant
acceleration are summarized in Table 5–2. One is useful for calculating
velocity and three are equations for position. When solving problems
involving constant acceleration, determine what information is given
and what is unknown, then choose the appropriate equation. These
equations, along with velocity-time and position-time graphs, provide
the mathematical models you need to solve motion problems.
Pocket Lab
Direction of
Acceleration
Tape a bubble level onto the top
of a laboratory cart. Center the
bubble. Observe the direction
of the motion of the bubble as
you pull the cart forward, move
it at constant speed, and allow
it to coast to a stop. Relate the
motion of the bubble to the
acceleration of the cart. Predict
what would happen if you tie
the string to the back of the
cart and repeat the experiment.
Try it.
Analyze and Conclude Draw
motion diagrams for the cart as
you moved it in the forward
direction and it coasted to a stop
and as you repeated the experiment in the opposite direction.
5.3 Acceleration
99
A car moving along a highway passes a
parked police car with a radar detector. Just
as the car passes, the police car starts to
pursue, moving with a constant acceleration.
The police car catches up with the car just as
it leaves the jurisdiction of the policeman.
Velocity
Problem
Time
Position
Ball and Car Race
Hypothesis
Time
Sketch the position-versus-time graphs and
the velocity-versus-time graphs for this
chase, then simulate the chase.
Possible Materials
battery-powered car
wood block
90-cm-long grooved track
1-in. steel ball
stopwatch
graph paper
Plan the Experiment
1. Identify the variables in this activity.
2. Determine how you will give the ball a
constant acceleration.
3. Devise a method to ensure that both
objects reach the end of the track at the
same time.
4. Construct a data table that will show the
positions of both objects at the beginning,
the halfway point, and the end of the chase.
5. Check the Plan Review your plan with
your teacher before you begin the race.
6. Construct p-t and v-t graphs for both
objects. Use technology to construct these
graphs if possible. Identify the relationships between variables.
7. Dispose of materials that cannot be
reused or recycled. Put away materials
that can be used again.
100
A Mathematical Model of Motion
Analyze and Conclude
1. Comparing and Contrasting Compare
the velocities of the cars at the beginning
and at the end of the chase. Write a verbal
description.
2. Using Graphs At any time during the
chase, did the cars ever have the same
velocity? If so, mark these points on the
graphs.
3. Comparing and Contrasting Compare
the average velocity of the police car to
that of the car.
4. Calculating Results Calculate the
average speed of each car.
Apply
1. Explain why it took the police car so long
to catch the car after it sped by.
2. Analyze and evaluate the plots of the
speeder’s motion. Infer from the plots
the speeder’s acceleration.
3. If the speeder accelerated at the exact
same rate of the police car at the moment
the speeder passed the police car, would
the police car ever catch the speeder?
Predict how your graphs would change.
4. Develop a CBL lab that plots the velocity
of a non-accelerated object and an accelerated object. Describe your graphs.
TABLE 5–2
Equations of Motion for Uniform Acceleration
Equation
Variables
v v0 at
d d0 + 1/2(v0 v)t
d d0 + v0t 1/2at2
v 2 v02 2a(d d0)
t
t
t
d
d
d
F.Y.I.
Initial Conditions
v a
v
a
v a
d0
d0
d0
A dragster tries to obtain
maximum acceleration over
a quarter-mile course. The
fastest time on record for
the quarter mile is 4.480 s.
The highest final speed on
record is 330.23 mph.
v0
v0
v0
v0
Example Problem
Finding Displacement Under Constant Acceleration
In Chapter 3, you completed the first step in solving the following
problem by sketching the situation and drawing the motion diagram.
Now you can add the mathematical model. A car starts at rest and
speeds up at 3.5 m/s2 after the traffic light turns green. How far will it
have gone when it is going 25 m/s?
Sketch the Problem
Begin
• Sketch the situation.
• Establish coordinate axes.
• Draw a motion diagram.
End
0
+x
v
Begin
Known:
End
a
Unknown:
d0 0.0 m
v 25 m/s
v0 0.0 m/s
a 3.5 m/s2
Strategy:
Refer to Table 5–2. Use an equation
containing v, a, and d.
d?
Calculations:
v2 v02 2a(d d0)
d d0 (v2 v02)/(2a)
0.0 m [(25 m/s)2 (0.0 m/s)2]/(23.5 m/s2)
89 m
• Is the unit correct? Dividing m2/s2 by m/s2 results in m, the correct unit for position.
• Does the sign make sense? It is positive, in agreement with both
the pictorial and physical models.
• Is the magnitude realistic? The displacement is almost the length
of a football field. It seems large, but 25 m/s is fast (about
55 mph), and the acceleration, as you will find in the next example problem, is not very great. Therefore, the result is reasonable.
5.3 Acceleration
101
Example Problem
Two-Part Motion
The driver of the car in the previous example problem, traveling
at a constant 25 m/s, sees a child suddenly run into the road. It
takes the driver 0.45 s to hit the brakes. As it slows, the car has a
steady acceleration of 8.5 m/s2. What’s the total distance the car
moves before it stops?
Begin
Sketch the Problem
Reacting
1
• Label your drawing with “begin” and “end.”
• Choose a coordinate system and create
the motion diagram.
• Use subscripts to distinguish the three
positions in the problem.
Known:
v2 25 m/s
d2 ?
v1 25 m/s
a23 8.5
d3 ?
a12 0.0
v3 0.0 m/s
m/s2
3
+x
0
Unknown:
d1 0.0 m
Braking
2
v
Begin
m/s2
End
a 12
End
a 23
t2 0.45 s
Strategy:
Calculations:
There are two parts to the problem: the interval
of reacting and the interval of braking.
Reacting: Find the distance the car travels. During this time, the velocity and time are known
and the velocity is constant.
Reacting: d2 vt
d2 (25 m/s)(0.45 s)
11 m
Braking: Find the distance the car moves while
braking. The initial and final velocities are
known. The acceleration is constant and negative, as shown in the motion diagram.
Braking: v32 v22 2a23(d3 d2)
The position of the car when the brakes are
applied, d2, is the solution of the first part of the
problem; it is needed to solve the second part.
d3 d2 (v32 v22)/(2a23)
0.0 (25 m/s)2
11
m
3
2(8.5 m/s2)
3 48 m
• Is the unit correct? Performing algebra on the units verifies the
distance in meters.
• Do the signs make sense? Both d3 and d2 are positive, as they
should be.
• Is the magnitude realistic? The braking distance is much smaller
than it was in the previous example problem, which makes sense
because the magnitude of the acceleration is larger.
102
A Mathematical Model of Motion
Practice Problems
For all problems, sketch the situation, assign variables, create a motion
diagram, and then develop a mathematical model.
27. A race car traveling at 44 m/s slows at a constant rate to a velocity
of 22 m/s over 11 s. How far does it move during this time?
28. A car accelerates at a constant rate from 15 m/s to 25 m/s while
it travels 125 m. How long does it take to achieve this speed?
29. A bike rider accelerates constantly to a velocity of 7.5 m/s during
4.5 s. The bike’s displacement during the acceleration was 19 m.
What was the initial velocity of the bike?
30. An airplane starts from rest and accelerates at a constant
3.00 m/s2 for 30.0 s before leaving the ground.
a. How far did it move?
b. How fast was it going when it took off?
5.3
Section Review
1a. Give an example of an object that is
slowing down but has a positive
acceleration.
b. Give an example of an object that is
speeding up but has a negative acceleration.
5. If you are given a table of velocities
of an object at various times, how
could you find out if the acceleration
was constant?
2a. If an object has zero acceleration,
does that mean its velocity is zero?
Give an example.
b. If an object has zero velocity at some
instant, does that mean its acceleration is zero? Give an example.
3. Is km/h/s a unit of acceleration?
Is this unit the same as km/s/h?
Explain.
4. Figure 5–17 is a strobe photo of
a horizontally moving ball. What
you need and what measurements
would you make to estimate the
acceleration?
FIGURE 5–17
6. If you are given initial and final
velocities and the constant acceleration of an object, and you are asked
to find the displacement, which
equation would you use?
7.
Critical Thinking Describe how you
could calculate the acceleration of an
automobile. Specify the measuring
instruments and the procedures you
would use.
5.3 Acceleration
103
5.4
Free Fall
D
OBJ ECTIVES
• Recognize the meaning of
the acceleration due to
gravity.
•
Define the magnitude of the
acceleration due to gravity
as a positive quantity and
determine the sign of the
acceleration relative to the
chosen coordinate system.
•
Use the motion equations
to solve problems involving
freely falling objects.
FIGURE 5–18 If the upward
direction is chosen as positive,
then both the velocity and the
acceleration of this apple in free
fall are negative.
104
rop a sheet of paper. Crumple it, then drop it again.
Its motion is different in the two instances. So is
the motion of a pebble falling through water compared
with the same pebble falling through air. Do heavier
objects fall faster than lighter ones? The answer depends
upon whether you drop sheets of paper or rocks.
Acceleration Due to Gravity
Galileo Galilei recognized about 400 years ago that to make progress
in the study of the motion of falling bodies, the effects of air or water,
the medium through which the object falls, had to be ignored. He also
knew that he had no means of recording the fall of objects, so he rolled
balls down inclined planes. By “diluting” gravity in this way, he could
make careful measurements even with simple instruments.
Galileo found that, neglecting the effect of the air, all freely falling
objects had the same acceleration. It didn’t matter what they were made
of, what their masses were, from how high they were dropped, or
whether they were dropped or thrown. The magnitude of the acceleration of falling objects is given a special symbol, g, equal to 9.80 m/s2.
We now know that there are small variations in g at different places on
Earth, and that 9.80 m/s2 is the average value.
Note that g is a positive quantity. You will never use a negative value
of g in a problem. But don’t things accelerate downward, and isn’t
down usually the negative direction? Although this is true, remember
that g is only the magnitude of the acceleration, not the acceleration
itself. If upward is defined to be the positive direction, then the acceleration due to gravity is equal to g. The acceleration due to gravity
is the acceleration of an object in free fall that results from the influence of Earth’s gravity. Suppose you drop a rock. One second later, its
velocity is 9.80 m/s downward. One second after that, its velocity is
19.60 m/s downward. For each second that the rock is falling, its downward velocity increases by 9.80 m/s.
Look at the strobe photo of a dropped apple in Figure 5–18. The
time interval between the photos is 1/120 s. The displacement between
each pair of images increases, so the speed is increasing. If the upward
direction is chosen as positive, then the velocity is becoming more and
more negative.
Could this photo be of a ball thrown upward? If you again choose
upward as the positive direction, then the ball leaves your hand with a
positive velocity of, say, 20.0 m/s. The acceleration is downward, so a is
negative. That is, a g 9.80 m/s2. This means that the speed of
the ball becomes less and less, which is in agreement with the strobe
A Mathematical Model of Motion
v (m/s)
v (m/s)
x (m)
20
0.5
25
0
0
t (s)
–20
0
– 0.5
4
a
x (m)
20.41
20.40
2
2.05
t (s)
2.1
0
b
t (s)
0
1
2
3
4
20.39
t (s)
2
2.05
2.1
d
c
photo. After 1 s, the ball’s velocity is reduced by 9.80 m/s, so it is now
traveling at 10.2 m/s. After 2 s, the velocity is 0.4 m/s, and the ball is
still moving upward. After the next second, the ball’s velocity, being
reduced by another 9.80 m/s, is now 9.4 m/s. The ball is now moving downward. After 4 s, the velocity is 19.2 m/s, meaning that it is
falling even faster. The velocity-time graph of the ball’s flight is shown
in Figure 5–19a. Figure 5–19b shows what happens at around 2 s,
where the velocity changes smoothly from positive to negative. At an
instant of time, near 2.04 s, the ball’s velocity is zero.
The position-time graphs in Figure 5–19c and d show how the ball’s
height changes. The ball has its maximum height when its velocity is zero.
FIGURE 5–19 In a coordinate
system in which the upward
direction is positive, the velocity
of the thrown ball decreases
until it becomes zero at 2.04 s,
then it increases in the negative
direction.
Free Fall
➥ Answers question from page 80.
Example Problem
The Demon Drop
The Demon Drop ride at Cedar Point Amusement
Park falls freely for 1.5 s after starting from rest.
a. What is its velocity at the end of 1.5 s?
b. How far does it fall?
+x
Sketch the Problem
0
• Choose a coordinate system with
a positive axis upward and the origin
at the initial position of the car.
• Label “begin” and “end.”
• Draw a motion diagram showing
that both a and v are downward
and, therefore, negative.
Known:
a g 9.80
d0 0
Begin
v
a
End
Unknown:
m/s2
d ?
v ?
v0 0
t 1.5 s
Continued on next page
5.4 Free Fall
105
Strategy:
Calculations:
a. Use the equation for velocity at constant
acceleration.
v v0 at
v 0 (9.80 m/s2)(1.5 s) 15 m/s
d d0 v0t 1/2at2
b. Use the equation for displacement when
time and constant acceleration are known.
d 0 0 1/2(9.80 m/s2)(1.5 s)2 11 m
• Are the units correct? Performing algebra on the units verifies
velocity in m/s and position in m.
• Do the signs make sense? Negative signs agree with the diagram.
• Are the magnitudes realistic? Yes, when judged by the photo at
the opening of this chapter in which the car is about the height
of a person, about 2 m.
Practice Problems
31. A brick is dropped from a high scaffold.
a. What is its velocity after 4.0 s?
b. How far does the brick fall during this time?
32. A tennis ball is thrown straight up with an initial speed of
22.5 m/s. It is caught at the same distance above ground.
a. How high does the ball rise?
b. How long does the ball remain in the air? (Hint: The
time to rise equals the time to fall. Can you show this?)
33. A spaceship far from any star or planet accelerates uniformly
from 65.0 m/s to 162.0 m/s in 10.0 s. How far does it move?
5.4
Section Review
1. Gravitational acceleration on Mars is
about 1/3 that on Earth. Suppose you
could throw a ball upward with the
same velocity on Mars as on Earth.
a. How would the ball’s maximum
height compare to that on Earth?
b. How would its flight time compare?
2. Research and describe Galileo’s contributions to physics.
106
A Mathematical Model of Motion
3.
Critical Thinking When a ball is thrown
vertically upward, it continues upward
until it reaches a certain position, then
it falls down again. At that highest
point, its velocity is instantaneously
zero. Is the ball accelerating at the
highest point? Devise an experiment
CHAPTER
5 REVIEW
Summary
Key Terms
5.1 Graphing Motion in
One Dimension
• The slope of the tan-
• Position-time graphs can be used to find
5.1
• uniform motion
5.3
• constant
acceleration
• instantaneous
acceleration
the velocity and position of an object,
and where and when two objects meet.
• A description of motion can be obtained
by interpreting graphs, and graphs can
be drawn from descriptions of motion.
• Equations that describe the position of
an object moving at constant velocity
can be written based on word and
graphical representations of problems.
•
•
•
5.4
• acceleration
due to gravity
5.2 Graphing Velocity in
One Dimension
• Instantaneous velocity is the slope of
the tangent to the curve on a positiontime graph.
• Velocity-time graphs can be used to
determine the velocity of an object and
the time when two objects have the
same velocity.
• The area under the curve on a velocitytime graph is displacement.
5.3 Acceleration
• The acceleration of an object is the slope
of the curve on a velocity-time graph.
•
•
gent to the curve
on a v-t graph is
the instantaneous
acceleration of the object.
Velocity-time graphs and motion diagrams can be used to find the sign of
the acceleration.
Both graphs and equations can be used
to find the velocity of an object undergoing constant acceleration.
Three different equations give the displacement of an object under constant
acceleration, depending on what quantities are known.
The mathematical model completes the
solution of motion problems.
Results obtained by solving a problem
must be tested to find out whether they
are reasonable.
5.4 Free Fall
• The magnitude of the acceleration due
to gravity (g 9.80 m/s2) is always a
positive quantity. The sign of acceleration depends upon the choice of the
coordinate system.
• Motion equations can be used to solve
problems involving freely falling objects.
Key Equations
5.1
d
d1 d0
v t
t1 t0
d d0 vt
5.3
v
v1 v0
a t
t t
v v0 at
1
0
d d0 v0t 1/2at2
v2 v02 2a(d – d0)
d d0 1/2(v v0)t
Reviewing Concepts
Section 5.1
1. A walker and a runner leave your front
door at the same time. They move in
the same direction at different constant
velocities. Describe the position-time
graphs of each.
Chapter 5 Review
107
CHAPTER 5 REVIEW
2. What does the slope of the tangent to the curve
on a position-time graph measure?
Section 5.2
3. If you know the positions of an object at two
points along its path, and you also know the
time it took to get from one point to the other,
can you determine the particle’s instantaneous
velocity? Its average velocity? Explain.
4. What quantity is represented by the area under
a velocity-time curve?
5. Figure 5–20 shows the velocity-time graph for
an automobile on a test track. Describe how
the velocity changes with time.
Section 5.4
12. Explain why an aluminum ball and a steel
ball of similar size and shape, dropped from
the same height, reach the ground at the
same time.
13. Give some examples of falling objects for which
air resistance cannot be ignored.
14. Give some examples of falling objects for
which air resistance can be ignored.
Applying Concepts
v (m/s)
30
25
20
15
10
5
00
11. Write a summary of the equations for position,
velocity, and time for an object experiencing
uniformly accelerated motion.
5
10
15
20
25
30
35
t (s)
15. Figure 5–20 shows the velocity-time graph of
an accelerating car. The three “notches” in the
curve occur where the driver changes gears.
a. Describe the changes in velocity and
acceleration of the car while in first gear.
b. Is the acceleration just before a gear change
larger or smaller than the acceleration just
16. Explain how you would walk to produce each
of the position-time graphs in Figure 5–21.
FIGURE 5–20
x
x
D
Section 5.3
6. What does the slope of the tangent to the curve
on a velocity-time graph measure?
7. A car is traveling on an interstate highway.
a. Can the car have a negative velocity and a
positive acceleration at the same time?
Explain.
b. Can the car’s velocity change signs while it is
traveling with constant acceleration?
Explain.
8. Can the velocity of an object change when its
acceleration is constant? If so, give an example.
If not, explain.
9. If the velocity-time curve is a straight line parallel to the t-axis, what can you say about the
acceleration?
10. If you are given a table of velocities of an
object at various times, how could you find out
if the acceleration of the object is constant?
108
A Mathematical Model of Motion
C
B
A
t
E
F
G
H
t
FIGURE 5–21
17. Use Figure 5–20 to determine during what
time interval the acceleration is largest and
during what time interval the acceleration
is smallest.
18. Solve the equation v v0 at for acceleration.
19. Figure 5–22 is a position-time graph of two
people running.
a. Describe the position of runner A relative to
runner B at the y-intercept.
b. Which runner is faster?
c. What occurs at point P and beyond?
CHAPTER 5 REVIEW
x
v
A
er
v
P
nn
a
Ru
nn
er
B
Ru
t
d (m)
12
A
B
6
0
2
b
t
FIGURE 5–24
FIGURE 5–22
0
t
4
6
8
10
t (s)
FIGURE 5–23
20. Figure 5–23 is a position-time graph of the
motion of two cars on a road.
a. At what time(s) does one car pass the other?
b. Which car is moving faster at 7.0 s?
c. At what time(s) do the cars have the same
velocity?
d.Over what time interval is car B speeding up
all the time?
e. Over what time interval is car B slowing
down all the time?
21. Look at Figure 5–24.
a. What kind of motion is represented by a?
b. What does the area under the curve
represent?
c. What kind of motion is represented by b?
d.What does the area under the curve
represent?
22. An object shot straight up rises for 7.0 s before
it reaches its maximum height. A second object
falling from rest takes 7.0 s to reach the
ground. Compare the displacements of the two
objects during this time interval.
23. Describe the changes in the velocity of a ball
thrown straight up into the air. Then describe
the changes in the ball’s acceleration.
24. The value of g on the moon is 1/6 of its value
on Earth.
a. Will a ball dropped by an astronaut hit the
surface of the moon with a smaller, equal, or
larger speed than that of a ball dropped from
the same height to Earth?
b. Will it take more, less, or equal time to fall?
25. Planet Dweeb has three times the gravitational
acceleration of Earth. A ball is thrown vertically
upward with the same initial velocity on Earth
and on Dweeb.
a. How does the maximum height reached by
the ball on Dweeb compare to the maximum
height on Earth?
b. If the ball on Dweeb were thrown with three
times greater initial velocity, how would that
26. Rock A is dropped from a cliff; rock B is
thrown upward from the same position.
a. When they reach the ground at the bottom
of the cliff, which rock has a greater velocity?
b. Which has a greater acceleration?
c. Which arrives first?
Problems
Section 5.1
27. Light from the sun reaches Earth in 8.3 min.
The velocity of light is 3.00 108 m/s. How far
is Earth from the sun?
28. You and a friend each drive 50.0 km. You travel
at 90 km/h; your friend travels at 95.0 km/h.
How long will your friend wait for you at the
end of the trip?
Chapter 5 Review
109
CHAPTER 5 REVIEW
29. The total distance a steel ball rolls down an
incline at various times is given in Table 5–3.
a. Draw a position-time graph of the motion of
the ball. When setting up the axes, use five
divisions for each 10 m of travel on the
d-axis. Use five divisions for 1 s of time on
the t-axis.
b. What type of curve is the line of the graph?
c. What distance has the ball rolled at the end
of 2.2 s?
TABLE 5–3
Distance versus Time
Time (s)
Distance (m)
0.0
1.0
2.0
3.0
4.0
5.0
0.0
2.0
8.0
18.0
32.9
50.0
30. A cyclist maintains a constant velocity of
5.0 m/s. At time t 0.0, the cyclist is
250 m from point A.
a. Plot a position-time graph of the cyclist’s
location from point A at 10.0-s intervals for
60.0 s.
b. What is the cyclist’s position from point A
at 60.0 s?
c. What is the displacement from the starting
position at 60.0 s?
31. From the position-time graph in Figure 5–25,
construct a table showing the average velocity
of the object during each 10-s interval over the
entire 100 s.
x (m)
500
400
300
200
100
0
t (s)
0
10
30
50
70
FIGURE 5–25
110
A Mathematical Model of Motion
90
32. Plot the data in Table 5–4 on a position-time
graph. Find the average velocity in the time
interval between 0.0 s and 5.0 s.
TABLE 5–4
Position versus Time
Position, d (m)
0.0
1.0
2.0
3.0
4.0
5.0
30
30
35
45
60
70
33. You drive a car for 2.0 h at 40 km/h, then for
another 2.0 h at 60 km/h.
a. What is your average velocity?
b. Do you get the same answer if you drive
1.0 102 km at each of the two speeds?
34. Use the position-time graph in Figure 5–25 to
find how far the object travels
a. between t 0 s and t 40 s.
b. between t 40 s and t 70 s.
c. between t 90 s and t 100 s.
35. Do this problem on a worksheet. Both car A
and car B leave school when a clock reads zero.
Car A travels at a constant 75 km/h, and car B
travels at a constant 85 km/h.
a. Draw a position-time graph showing the
motion of both cars.
b. How far are the two cars from school when
the clock reads 2.0 h? Calculate the distances
using the equation for motion and show
c. Both cars passed a gas station 120 km from
the school. When did each car pass the gas
station? Calculate the times and show them
36. Draw a position-time graph for two cars driving to the beach, which is 50 km from school.
At noon Car A leaves a store 10 km closer to
the beach than the school is and drives at
40 km/h. Car B starts from school at 12:30 P.M.
and drives at 100 km/h. When does each car
get to the beach?
CHAPTER 5 REVIEW
37. Two cars travel along a straight road. When a
stopwatch reads t 0.00 h, car A is at dA 48.0 km moving at a constant 36.0 km/h.
Later, when the watch reads t 0.50 h, car B is
at dB 0.00 km moving at 48.0 km/h. Answer
the following questions, first, graphically by
creating a position-time graph, and second,
algebraically by writing down equations for the
positions dA and dB as a function of the stopwatch time, t.
a. What will the watch read when car B passes
car A?
b. At what position will car B pass car A?
c. When the cars pass, how long will it have
been since car A was at the reference point?
38. A car is moving down a street at 55 km/h. A
child suddenly runs into the street. If it takes
the driver 0.75 s to react and apply the brakes,
how many meters will the car have moved
before it begins to slow down?
Section 5.2
39. Refer to Figure 5–23 to find the instantaneous
speed for
a. car B at 2.0 s.
b. car B at 9.0 s.
c. car A at 2.0 s.
40. Refer to Figure 5–26 to find the distance the
moving object travels between
a. t 0 s and t 5 s.
b. t 5 s and t 10 s.
c. t 10 s and t 15 s.
d.t 0 s and t 25 s.
v (m/s)
30
20
10
0
t (s)
0
5
10
15
20
25
30
FIGURE 5–26
41. Find the instantaneous speed of the car in
Figure 5–20 at 15 s.
42. You ride your bike for 1.5 h at an average
velocity of 10 km/h, then for 30 min at
15 km/h. What is your average velocity?
43. Plot a velocity-time graph using the information in Table 5–5, then answer the questions.
a. During what time interval is the object
speeding up? Slowing down?
b. At what time does the object reverse
direction?
c. How does the average acceleration of the
object in the interval between 0 s and 2 s
differ from the average acceleration in the
interval between 7 s and 12 s?
TABLE 5–5
Velocity versus Time
Time
(s)
Velocity
(m/s)
Time
(s)
Velocity
(m/s)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
4.0
8.0
12.0
14.0
16.0
16.0
14.0
7.0
8.0
9.0
10.0
11.0
12.0
12.0
8.0
4.0
0.0
4.0
8.0
Section 5.3
44. Find the uniform acceleration that causes a
car’s velocity to change from 32 m/s to 96 m/s
in an 8.0-s period.
45. Use Figure 5–26 to find the acceleration of the
moving object
a. during the first 5 s of travel.
b. between the fifth and the tenth second of
travel.
c. between the tenth and the 15th second of travel.
d. between the 20th and 25th second of travel.
46. A car with a velocity of 22 m/s is accelerated
uniformly at the rate of 1.6 m/s2 for 6.8 s.
What is its final velocity?
47. A supersonic jet flying at 145 m/s is accelerated
uniformly at the rate of 23.1 m/s2 for 20.0 s.
a. What is its final velocity?
b. The speed of sound in air is 331 m/s. How
many times the speed of sound is the plane’s
final speed?
48. Determine the final velocity of a proton that has
an initial velocity of 2.35 105 m/s, and then is
accelerated uniformly in an electric field at the
rate of 1.10 1012 m/s2 for 1.50 107 s.
Chapter 5 Review
111
CHAPTER 5 REVIEW
49. Determine the displacement of a plane that is
uniformly accelerated from 66 m/s to 88 m/s
in 12 s.
50. How far does a plane fly in 15 s while its
velocity is changing from 145 m/s to 75 m/s
at a uniform rate of acceleration?
51. A car moves at 12 m/s and coasts up a hill with
a uniform acceleration of 1.6 m/s2.
a. How far has it traveled after 6.0 s?
b. How far has it gone after 9.0 s?
52. A plane travels 5.0 102 m while being accelerated uniformly from rest at the rate of
5.0 m/s2. What final velocity does it attain?
53. A race car can be slowed with a constant acceleration of 11 m/s2.
a. If the car is going 55 m/s, how many meters
will it take to stop?
b. How many meters will it take to stop a car
going twice as fast?
54. An engineer must design a runway to accommodate airplanes that must reach a ground
velocity of 61 m/s before they can take off.
These planes are capable of being accelerated
uniformly at the rate of 2.5 m/s2.
a. How long will it take the planes to reach
takeoff speed?
b. What must be the minimum length of the
runway?
55. Engineers are developing new types of guns
that might someday be used to launch satellites
as if they were bullets. One such gun can give a
small object a velocity of 3.5 km/s, moving it
through only 2.0 cm.
a. What acceleration does the gun give this
object?
b. Over what time interval does the acceleration
take place?
56. Highway safety engineers build soft barriers so
that cars hitting them will slow down at a safe
rate. A person wearing a seat belt can withstand
an acceleration of 3.0 102 m/s2. How thick
should barriers be to safely stop a car that hits
a barrier at 110 km/h?
57. A baseball pitcher throws a fastball at a speed
of 44 m/s. The acceleration occurs as the
pitcher holds the ball in his hand and moves it
through an almost straight-line distance of
3.5 m. Calculate the acceleration, assuming it
112
A Mathematical Model of Motion
is uniform. Compare this acceleration to the
acceleration due to gravity, 9.80 m/s2.
58. Rocket-powered sleds are used to test the
responses of humans to acceleration. Starting
from rest, one sled can reach a speed of
444 m/s in 1.80 s and can be brought to a
stop again in 2.15 s.
a. Calculate the acceleration of the sled when
starting, and compare it to the magnitude of
the acceleration due to gravity, 9.80 m/s2.
b. Find the acceleration of the sled when
braking and compare it to the magnitude of
the acceleration due to gravity.
59. Draw a velocity-time graph for each of the
graphs in Figure 5–27.
x
x
t
x
t
t
FIGURE 5–27
60. The velocity of an automobile changes over an
8.0-s time period as shown in Table 5–6.
a. Plot the velocity-time graph of the motion.
b. Determine the displacement of the car
during the first 2.0 s.
c. What displacement does the car have during
the first 4.0 s?
d.What displacement does the car have during
the entire 8.0 s?
e. Find the slope of the line between t 0.0 s
and t 4.0 s. What does this slope
represent?
f. Find the slope of the line between t 5.0 s
and t 7.0 s. What does this slope indicate?
TABLE 5–6
Velocity versus Time
Time
(s)
0.0
1.0
2.0
3.0
4.0
Velocity
(m/s)
0.0
4.0
8.0
12.0
16.0
Time
(s)
5.0
6.0
7.0
8.0
Velocity
(m/s)
20.0
20.0
20.0
20.0
Displacement (cm)
10
TABLE 5–7
Initial Velocity versus
Braking Distance
5
0
–5
Initial Velocity
(m/s)
Braking Distance
(m)
11
15
20
25
29
10
20
34
50
70
0
Velocity (m/s)
Fist
CHAPTER 5 REVIEW
–5
– 10
– 15
5
10
15
20
Time (ms)
25
30
35
FIGURE 5-28
61. Figure 5–28 shows the position-time and
velocity-time graphs of a karate expert’s fist as
it breaks a wooden board.
a. Use the velocity-time graph to describe
the motion of the expert’s fist during the
first 10 ms.
b. Estimate the slope of the velocity-time graph
to determine the acceleration of the fist
when it suddenly stops.
c. Express the acceleration as a multiple of the
gravitational acceleration, g = 9.80 m/s2.
d.Determine the area under the velocity-time
curve to find the displacement of the fist in
the first 6 ms. Compare this with the
position-time graph.
62. The driver of a car going 90.0 km/h suddenly
sees the lights of a barrier 40.0 m ahead. It takes
the driver 0.75 s to apply the brakes, and the
average acceleration during braking is 10.0 m/s2.
a. Determine whether the car hits the barrier.
b. What is the maximum speed at which the car
could be moving and not hit the barrier
40.0 m ahead? Assume that the acceleration
rate doesn’t change.
63. The data in Table 5–7, taken from a driver’s
handbook, show the distance a car travels
when it brakes to a halt from a specific
initial velocity.
a. Plot the braking distance versus the initial
velocity. Describe the shape of the curve.
b. Plot the braking distance versus the square of
the initial velocity. Describe the shape of
the curve.
c. Calculate the slope of your graph from
part b. Find the value and units of the
quantity 1/slope.
d.Does this curve agree with the equation
v02 2ad? What is the value of a?
64. As a traffic light turns green, a waiting car
starts with a constant acceleration of 6.0 m/s2.
At the instant the car begins to accelerate, a
truck with a constant velocity of 21 m/s passes
in the next lane.
a. How far will the car travel before it overtakes
the truck?
b. How fast will the car be traveling when it
overtakes the truck?
65. Use the information given in problem 64.
a. Draw velocity-time and position-time graphs
for the car and truck.
b. Do the graphs confirm the answer you
calculated for problem 64?
Section 5.4
66. An astronaut drops a feather from 1.2 m above
the surface of the moon. If the acceleration of
gravity on the moon is 1.62 m/s2 downward,
how long does it take the feather to hit the
moon’s surface?
67. A stone falls freely from rest for 8.0 s.
a. Calculate the stone’s velocity after 8.0 s.
b. What is the stone’s displacement during
this time?
68. A student drops a penny from the top of a
tower and decides that she will establish
a coordinate system in which the direction of
the penny’s motion is positive. What is the
sign of the acceleration of the penny?
Chapter 5 Review
113
CHAPTER 5 REVIEW
69. A bag is dropped from a hovering helicopter.
When the bag has fallen 2.0 s,
a. what is the bag’s velocity?
b. how far has the bag fallen?
70. A weather balloon is floating at a constant
height above Earth when it releases a pack
of instruments.
a. If the pack hits the ground with a velocity
of 73.5 m/s, how far did the pack fall?
b. How long did it take for the pack to fall?
71. During a baseball game, a batter hits a high
pop-up. If the ball remains in the air for 6.0 s,
how high does it rise? Hint: Calculate the
height using the second half of the trajectory.
72. Table 5–8 gives the positions and velocities of
a ball at the end of each second for the first
5.0 s of free fall from rest.
a. Use the data to plot a velocity-time graph.
b. Use the data in the table to plot a positiontime graph.
c. Find the slope of the curve at the end of
2.0 s and 4.0 s on the position-time graph.
Do the values agree with the table of
velocity?
d.Use the data in the table to plot a
position-versus-time-squared graph. What
type of curve is obtained?
e. Find the slope of the line at any point.
Explain the significance of the value.
f. Does this curve agree with the equation
d 1/2 gt2?
TABLE 5–8
Position and Velocity in Free Fall
Time (s)
0.0
1.0
2.0
3.0
4.0
5.0
Position (m)
0.0
4.9
19.6
44.1
78.4
122.5
Velocity (m/s)
0.0
9.8
19.6
29.4
39.2
49.0
73. The same helicopter in problem 69 is rising at
5.0 m/s when the bag is dropped. After 2.0 s,
a. what is the bag’s velocity?
114
A Mathematical Model of Motion
b. how far has the bag fallen?
c. how far below the helicopter is the bag?
74. The helicopter in problems 69 and 73 now
descends at 5.0 m/s as the bag is released.
After 2.0 s,
a. what is the bag’s velocity?
b. how far has the bag fallen?
c. how far below the helicopter is the bag?
75. What is common to the answers to problems
69, 73, and 74?
76. A tennis ball is dropped from 1.20 m above the
ground. It rebounds to a height of 1.00 m.
a. With what velocity does it hit the ground?
b. With what velocity does it leave the ground?
c. If the tennis ball were in contact with the
ground for 0.010 s, find its acceleration
while touching the ground. Compare the
acceleration to g.
Extra Practice For more
practice solving problems, go
to Extra Practice Problems,
Appendix B.
Critical Thinking Problems
77. An express train, traveling at 36.0 m/s, is accidentally sidetracked onto a local train track.
The express engineer spots a local train exactly
1.00 102 m ahead on the same track and
traveling in the same direction. The local engineer is unaware of the situation. The express
engineer jams on the brakes and slows the
express at a constant rate of 3.00 m/s2. If the
speed of the local train is 11.0 m/s, will the
express train be able to stop in time or will
there be a collision? To solve this problem,
take the position of the express train when it
first sights the local train as a point of origin.
Next, keeping in mind that the local train has
exactly a 1.00 102 m lead, calculate how far
each train is from the origin at the end of the
12.0 s it would take the express train to stop.
CHAPTER 5 REVIEW
a. On the basis of your calculations, would
you conclude that a collision will occur?
b. The calculations you made do not allow for
the possibility that a collision might take
place before the end of the 12 s required for
the express train to come to a halt. To check
this, take the position of the express train
when it first sights the local train as the
point of origin and calculate the position of
each train at the end of each second after
sighting. Make a table showing the distance
of each train from the origin at the end of
each second. Plot these positions on the
same graph and draw two lines. Use your
78. Which has the greater acceleration: a car that
increases its speed from 50 to 60 km/h, or a
bike that goes from 0 to 10 km/h in the same
time? Explain.
79. You plan a car trip on which you want to average 90 km/h. You cover the first half of the distance at an average speed of only 48 km/h.
What must your average speed be in the second
half of the trip to meet your goal? Is this reasonable? Note that the velocities are based on
half the distance, not half the time.
Going Further
Use a graphing calculator to fit a line to a positiontime graph of the data and to plot this line. Be sure to
set the display range of the graph so that all the data
fit on it. Find the slope of the line. What was the speed
of the car?
Applying CBLs Design a lab to measure the dis-
tance an accelerated object moves over time. Use
equal time intervals so that you can plot velocity
over time as well as distance. A pulley at the edge
of a table with a mass attached is a good way to
achieve uniform acceleration. Suggested materials
include a motion detector, CBL, lab cart, string,
pulley, C-clamp, and mass. Generate graphs of distance versus time and velocity versus time using
different masses on the pulley. How did the
change in mass affect your graphs?
Applying Calculators Members of a physics
class stood 25 m apart and used stopwatches
to measure the time a car driving down the
highway passed each person. The data they
compiled are shown in Table 5–9.
TABLE 5–9
Position versus Time
Time
(s)
0.0
1.3
2.7
3.6
5.1
Position
(m)
0.0
25.0
50.0
75.0
100.0
Time
(s)
5.9
7.0
8.6
10.3
Position
(m)
125.0
150.0
175.0
200.0
PHYSICS
To review content, do the
interactive quizzes on the
Glencoe Science Web site at
science.glencoe.com
Chapter 5 Review
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