# Chapter 2 Kinematics in One Dimension Newtonian mechanics ```July-15-14
10:39 AM
Chapter 2 Kinematics in One Dimension
Newtonian mechanics: kinematics and dynamics
Kinematics: mathematical description of motion (Ch 2, Ch 3)
Dynamics: how forces affect motion (Ch 4)
Summary of Key Concepts
• unit conversions (1.3)
• dimensional analysis (1.3)
• vector and scalar quantities (1.5 − 1.8)
• position, displacement, distance (2.1)
• velocity, speed, average velocity, and average speed (2.2)
• acceleration and average acceleration (2.3)
• kinematics equations for constant acceleration (2.4, 2.5)
• freely falling objects (2.6)
• graphical analysis of one-dimensional kinematics (2.7)
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Vector and scalar quantities
Some of the quantities that we use to describe motion
mathematically have magnitudes and directions; these are
called vector quantities. Some of the quantities that we use
to describe motion mathematically have only magnitudes,
with no directionality; these are called scalar quantities.
We'll get into the details of vectors and scalars later this
week, on Friday. In the meantime, every time a new
quantity is introduced, ask yourself if it is a vector or a
scalar. Does it have just a magnitude, or does it have a
directionality too?
In our textbook, vectors are indicated by boldface type and
by placing a small arrow above the symbol. (Other books
use boldface without the arrow, so watch for this in you
linear algebra class or in advanced courses.) I can't do
boldface very well in these notes, so I'll just place a small
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boldface very well in these notes, so I'll just place a small
arrow above a symbol to indicate a vector quantity. Scalar
quantities are indicated by symbols in italic type with no
arrow above the symbol.
The reason we postpone a detailed discussion of vectors
until the end of the week is that for motion in a straight line
(the main focus of this chapter) there are only two
directions, and so we can manage just by using a positive
sign for one of the directions and a negative sign for the
other direction, without having to use the full power of
vector notation. In the next chapter, when we discuss
motion in two dimensions, we'll need to use the full power
of vector notation, so we'll discuss vectors just before
they're needed.
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What is displacement?
Example: An object moves along a straight line with initial
position 9 m and final position 2 m. Determine the displacement.
Solution:
The displacement can be represented by the blue arrow in the
figure above. Note that the arrow points to the left, which
corresponds to the fact that the displacement is negative. For a
positive displacement, the arrow would point to the right.
If the line of motion is oriented differently (north-south, for
example, or in some other direction), then it's up to you to
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example, or in some other direction), then it's up to you to
choose a positive direction along the line; the other direction is
therefore negative.
How does distance differ from displacement?
Example: An object moves along a straight line from position 9 m
to position 2 m. Then the object moves along the same line
another 4 m to the right.
(a) Determine the total displacement.
(b) Determine the total distance travelled.
Solution:
How are average velocity and average speed calculated?
Example: An object moves along a straight line from position 7 m
to position 4 m in 3 s. Then the object moves along the same line
back to its starting position in 2 s.
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back to its starting position in 2 s.
(a) Determine the average velocity for the whole trip.
(b) Determine the average speed for the whole trip.
Solution:
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Example: An object moves along a straight line from position 8 m
to position 3 m in 4 s. Then the object moves along the same line
another 2 m to the right in 1 s.
(a) Determine the average velocity for each leg of the trip, and
for the whole trip.
(b) Determine the average speed for each leg of the trip, and for
the whole trip.
Solution:
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The average velocity can be interpreted by pretending the actual
motion is replaced by the wiggly blue arrow in the diagram above.
The wiggly blue arrow represents the total displacement for the
entire journey. (At least it would if we replaced it by a straight
arrow; I made it wiggly just to distinguish it from the displacement
arrows of the actual motion.)
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The average speed is always the total distance divided by the
total time. For a straight line segment of a journey, with no
changes of direction, the average speed can also be calculated as
the length of the average velocity vector.
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Example: Alice jogs one complete lap around an oval track, a
distance of 200 m, in a time of 40 s.
(a) Determine Alice's displacement and average velocity.
(b) Determine Alice's distance travelled and her average speed.
Solution: (a)
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This example illustrates that the concepts of motion in one
dimension (i.e., straight-line motion) can also be successfully
used in certain cases for other types of motion. If the road is
curvy, but all we care about is scalar quantities such as
distances travelled and speeds, then using concepts of
straight-line motion work out fine. And for round-trip
motions (even if not in a straight line), the total displacement
is zero, so being sloppy and treating the zero vector as if it's
the zero scalar is acceptable. (Don't tell your math prof,
though!)
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Example: Basil drives from St. Catharines to Toronto, a distance
of 110 km, and then back again. His average speed on the way to
Toronto is 60 km/h, and his average speed on the way back is 90
km/h. Determine Basil's average speed for the whole trip.
Solution: Use the basic idea that average speed is distance
divided by time:
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Question: Does the distance between the two cities matter in
the previous example? Can you solve the problem if the distance
between the two cities is not given?
Solution: Yes, we can!
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Let's check this result using the values of the previous problem; sure
enough, if you substitute speeds of 60 km/h and 90 km/h into the result,
you reproduce the result of the previous problem. This is comforting.
Another way to play with this result is to play with extreme values. If the
speed to Toronto is 1 km/h and the speed on the return trip is 100
km/h, we expect that the average speed for the entire trip is a lot closer
to 1 km/h than to 100 km/h. Sure enough, if you substitute these values
into the previous formula, you obtain about 2 km/h. This is a powerful
reminder that to determine the average speed in situations such as this
one, you can't just add the two speeds and divide by 2.
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Unit conversions
We'll take a break from kinematics to discuss how to convert from one
unit to another. I assume that you know all about the SI system of units,
which is discussed in high school. If you need a refresher, here is a link:
http://physics.nist.gov/cuu/Units/
Also, check Section 1.3 in the textbook.
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Also, check Section 1.3 in the textbook.
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Example: Convert 40 km/h to m/s.
Solution 1: Note that 1 km is equivalent to 1000 m, and that 1 h is
equivalent to 3600 s. Thus,
Solution 2: Because 1000 m is equivalent to 1 km, their ratio is
exactly 1:
Therefore, one can multiply by either of these quotients to
convert units. Similarly,
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Thus, to convert 40 km/h to m/s, you can proceed as
follows:
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Example: Convert 8.7 m/s to km/h.
Solution 1:
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Solution 2:
Dimensional analysis
Consider two objects, which have lengths of 5 m and 2 km,
respectively. Consider their lengths as physical quantities in their
own right, abstracted away from the objects themselves. The
two lengths are different, sure, but nevertheless there is
something the same about them, because they are both lengths.
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something the same about them, because they are both lengths.
For example, if the first object has a mass of 20 kg, then its
length 5 m and its mass 20 kg are also different, but they are
different kinds of quantities.
To help us analyze the distinctions discussed in the previous
paragraph, we introduce the concept of the dimension of a
physical quantity. Although the quantities 5 m and 2 km are
different quantities, and they also have different units, we say
that they have the same dimension, because they are both
lengths.
However, the quantities 5 m and 20 kg have different
dimensions, because one is a length and one is a mass.
The basic dimensions are mass (M), length (L), time (T), and a
few others that we won't bother about until future courses.
A fundamental principle of physics is that:
Each term in a valid physical equation must have the same
dimension.
There is an every-day slogan that corresponds to this
fundamental principle: "You can't add apples and oranges,"
because it's meaningless to do so. Sure, someone can give you 5
apples and 3 oranges, and sure that means that you have been
given 8 items, but in physics
3 kg + 4 m
has no meaning.
Now, maybe someday someone creative (maybe you?) may
figure out a way to make sense of 3 kg + 4 m, but right now this
doesn't make sense to us, so we'll stick with the principle that
every term of a valid physical equation must have the same
dimension.
Even though you can't meaningfully add or subtract quantities
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Even though you can't meaningfully add or subtract quantities
with different dimensions, you can multiply or divide quantities
with different dimensions to obtain different meaningful ones.
We'll see this throughout the course; for example, a quantity
with units kg.m2/s2 might represent energy, or maybe something
else.
A similar physical principle is that in a valid physical equation the
argument of a non-polynomial function must be dimensionless;
i.e., a number that has no units. For example, exponents must be
dimensionless, angles substituted into sine, cosine, etc.,
functions must be dimensionless, quantities that you take the
logarithm of must be dimensionless, etc.
An example of dimensional analysis
The fact that each term of a valid physical equation must have
the same dimension gives us a tool for (partially) checking
physical formulas, and also for guessing them; this tool is called
dimensional analysis, and we'll discuss an example shortly.
Dimensional analysis provides only a partial check, because
method, as we'll see.
For example, one of the kinematics equations that we'll use later
this chapter is
or, if you wish,
Check the dimensions of each term, and you'll see that they are
the same, so this is a partial check on the equation:
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Each term in the equation has dimensions of length, so the
equation has the possibility of being correct. To be certain that
it's correct, we would have to carefully derive it using physical
principles, correct mathematics, and correct logic. And of
course, the equation would have to then be extensively tested
by observations and experiments; in science, experimentation
and observation are the ultimate arbiters.
The reason that dimensional analysis is only a partial check is
that had we written an incorrect version of the above
kinematics equation that had incorrect dimensionless
coefficients, such as
Dimensional analysis could never detect the error.
Dimensional analysis only serves to test whether the principle
that all terms in a valid physical equation have the same
dimension has been satisfied. Dimensional analysis does not
verify that the equation is correct!
Nevertheless, dimensional analysis is a beloved and useful
tool for physicists. It can also be used to guess the form of a
physical equation. A classic example is the simple pendulum
that all you 1P91 students will be studying in Experiment 1.
What factors might affect the period of a simple pendulum?
Well, hang a weight on the end of a string and play with it,
and you'll soon come up with several factors that seem to
play a role. We don't have a string handy, but we might guess
pendulum that might change its period.
Guess using dimensional analysis:
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Guess using dimensional analysis:
Remember, the guess using dimensional analysis is only a
guess. To determine the correct formula, one would have to
analyze the situation using the relevant dynamical theory
(Newton's laws of motion, in this case). Then the formula
would have to be extensively tested experimentally. All you
1P91 students will do some of this work in Experiment 1.
If you're interested, you can study the development in Section
10.4 of the textbook, where the following correct formula is
derived:
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Note that dimensional analysis was successful in providing the
right functional form, but it is not possible to determine the
correct numerical factors without a full analysis based on
Newton's laws of motion. But just the functional form is often
enough to tell you what you want to know; for example,
knowing that the period is proportional to the square root of
the length of the pendulum gives us the useful information
that increasing the length of the pendulum increases the
period. The fact that mass does not appear in the formula
tells us that the period is independent of the mass.
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What is acceleration? How is it calculated?
Example: Alice drives at a speed of 50 km/h, then enters a
highway, and so increases her speed to 100 km/h in a time of 8 s.
Determine her average acceleration, in both (km/h)/s and in
m/s2.
Solution:
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Typically, it is easier for us humans to understand units such as
(km/h)/s, because of our experience driving cars. However, it is
often easier to solve problems using the standard unit of
acceleration, which is m/s2. This means being able to quickly and
accurately go from one unit to the other is useful; it's a skill worth
practicing.
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Example: Alice now leaves the highway, and so decreases her
speed from 100 km/h to 60 km/h in a time of 10 s. Determine
her acceleration, in both (km/h)/s and in m/s2.
Solution:
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Is acceleration a vector or a scalar?
Page 53, Problem 19 The initial velocity and acceleration of four
moving objects at a given instant in time are given in the
following table. Determine the final speed of each of the objects,
assuming that the time elapsed since t = 0 s is 2.0 s.
Initial
Acceleration
Velocity v0 a
(a) +12 m/s +3.0 m/s2
(b) +12 m/s −3.0 m/s2
(c.) −12 m/s +3.0 m/s2
(d) −12 m/s −3.0 m/s2
Solution:
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The conclusion we draw from this example is that (for motion in a
straight line) if the acceleration is in the same direction as the
velocity (i.e., they have the same signs) then the object speeds up,
whereas if the acceleration opposes the velocity (i.e., they have
opposite signs) then the object slows down.
In the previous paragraph we have been speaking about
acceleration as a vector. In every-day language, we often treat
acceleration as a scalar; that is, we use "acceleration" to mean
"speeding up" and "deceleration" to mean "slowing down." It
would have been so much easier to understand and communicate
these concepts if we had two separate words, one for acceleration
as a vector and a different one for acceleration as a scalar (just as
we use velocity and speed), but we only have one word, so that's
life.
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Page 53, Problem 23 Two motorcycles are travelling east with
different speeds. However, four seconds later, they have the same
speed. During this four-second interval, cycle A has an average
acceleration of 2.0 m/s2 east, while cycle B has an average acceleration
of 4.0 m/s2 east. By how much did the speeds differ at the beginning of
the four-second interval, and which motorcycle was moving faster?
Discussion: There are two moving objects here, so we should be
careful to label the kinematical variables differently for each object.
Solution:
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We are not asked anything about the positions, and we are not given any
information about the positions; perhaps we can just ignore positions? If
that strategy is to work, we would need to find a way to connect the
desired quantities (initial velocities) with the given quantities
(accelerations and times). But this is indeed possible:
Note that the solution can also be obtained in a relatively simple
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Note that the solution can also be obtained in a relatively simple
way by drawing a velocity-time graph. Study the following figure
and see if you can understand how to use it to solve the
problem:
Does the solution make sense? Every second, motorcycle B gains on
motorcycle A by 2 m/s. Therefore, after 4 s, motorcycle B gains on
motorcycle A by 8 m/s. But after 4 s they have the same velocity. This
means that at t = 0, A must have been going faster than B by 8 m/s.
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Understanding velocity-time graphs
The slope of a velocity-time graph represents acceleration.
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The area "under" a velocity-time graph represents
displacement. The first diagram below illustrates this fact
for a motion with constant velocity (i.e., zero acceleration),
and the second diagram below illustrates this fact for nonzero acceleration. (In both cases the acceleration is
constant.)
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Arguments based on the diagrams above lead us to basic
equations relating the kinematical variables that are valid for
constant acceleration. The following four equations are
sufficient for solving all the problems in the textbook.
(Formal derivations of the equations, for those who are
interested, appear later in the notes for this chapter.)
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What are the kinematics equations for constant acceleration?
How do you use them?
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Page 53, Problem 28 (a) Determine the magnitude of the average
acceleration of a skier who, starting from rest, reaches a speed of
8.0 m/s when going down a slope for 5.0 s.
(b) Determine how far the skier travels in this time.
Solution:
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Thus, the magnitude of the average acceleration is 1.6 m/s2.
(b) There are several possible methods, all illustrated below:
The displacement of the skier is 20 m.
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Page 53, Problem 29 A jetliner, traveling northward, is
landing with a speed of 69 m/s. Once the jet touches down, it
has 750 m of runway in which to reduce its speed to 6.1 m/s.
Determine the average acceleration (magnitude and
direction) of the plane during landing.
Solution:
One way to solve the problem is to begin with the velocitytime graph. Note that we could calculate the slope of the
graph (which is the required acceleration) if we knew the
time over which the acceleration takes place. We can do this
by writing an expression for the shaded area and equating it
to the given displacement, 750 m.
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Now that we know the time, we can determine the acceleration.
Thus, the magnitude of the acceleration is 3.15 m/s2, and the
acceleration is in the negative direction (that is, the acceleration is
in the south direction).
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An alternative solution:
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Example: Alice starts from rest and accelerates at a constant rate
of 2 m/s2 for 5 s. Then her acceleration suddenly increases to a
new constant rate of 4 m/s2, which she maintains for an additional
10 s.
(a) Determine her average acceleration for the entire time period.
(b) Determine her displacement during the entire time period.
Solution:
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Page 53, Problem 24 A basketball player starts from rest and
sprints to a speed of 6.0 m/s in 1.5 s. Assuming that the player
accelerates uniformly, determine the distance she runs.
Discussion: We assume that the player moves in a straight line
without changing direction along the line. In this case the
distance travelled is the same as the magnitude of the
displacement.
Solution 1:
Solution 2:
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Solution 2:
Solution 3:
Which solution do you prefer? All are good, so use the method
that is best for you. (However, the best problem-solvers are those
that are comfortable with many methods, so it’s worth practicing
a variety of different methods.
Solution 4:
Page 53, Problem 32 Two rockets are flying in the same
direction and are side by side at the instant their retrorockets
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direction and are side by side at the instant their retrorockets
fire. Rocket A has an initial velocity of +5800 m/s, whereas rocket
B has an initial velocity of +8600 m/s. After a time t both rockets
are again side by side, the displacement of each being zero. The
acceleration of rocket A is −15 m/s2. Determine the acceleration
of rocket B.
Discussion: Just as in the motorcycle problem solved earlier,
there are two moving objects here, so we should be careful to
label separate kinematical variables for each object:
Solution 1:
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Now substitute the expression for t on the left into the equation
on the right and solve for the acceleration of Rocket B:
Solution 2:
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Page 53, Problem 33 A car is travelling at 20.0 m/s, and
the driver sees a traffic light turn red. After 0.530 s (the
reaction time), the driver applies the brakes and the car
decelerates at 7.00 m/s2. Determine the stopping distance
of the car, as measured from the point where the driver
first sees the red light.
Solution:
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The car's stopping distance is 39.2 m.
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Page 54, Problem 34 A race driver has made a pit stop to refuel.
After refuelling, he starts from rest and leaves the pit area with an
acceleration of magnitude 6 m/s2; after 4.0 s, he enters the main
speedway. At the same instant, another car on the speedway
travelling at a constant speed of 70.0 m/s overtakes and passes the
entering car. The entering car maintains its acceleration. Determine
the time needed for the entering car to catch the other car.
Discussion: There are two moving objects, so use A and B to
distinguish their kinematical variables. For the first race car, call its
"initial" speed the speed it has after it accelerates for 4 s, which is
Solution 1:
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Yes, the cars are side-by-side at t = 0 s, but Car A overtakes Car B
after 15.3 s.
Solution 2:
Page 54, Problem 40 An airplane has a length of 59.7 m. The
runway on which the plane lands intersects another runway. The
width of the intersection is 25.0 m. The plane decelerates
through the intersection at a rate of 5.70 m/s2 and clears it with a
final speed of 45.0 m/s. Determine the time needed for the plane
to clear the intersection.
Discussion: I interpret "clearing the intersection" to mean that
the displacement of the plane is 59.7 m + 25.0 m = 84.7 m.
Let t = 0 represent the time at which the plane enters the
intersection; thus, we are looking for a positive value of t as the
solution of this problem (negative values of t represent times
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solution of this problem (negative values of t represent times
before the plane reaches the intersection, and so are not relevant
for this problem).
If we knew the value of vo instead of the value of v, we could just use the
third kinematics equation to solve the problem. But this would work if
we could express the initial velocity in terms of the final velocity, and
then substitute the resulting expression into the third kinematics
equation. Let`s try this.
Solution:
This is a quadratic equation with only one unknown, t, and this is
the variable that we seek. Thus, use the quadratic formula to
solve for t.
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Remember that the negative solution is not relevant for this
problem.
Page 54, Problem 42 A train is 92 m long and starts from rest
with a constant acceleration at time t = 0 s. At this instant, a car
just reaches the end of the train. The car is moving with a
constant speed. At a time t = 14 s, the car just reaches the front
of the train. Ultimately, however, the train pulls ahead of the car,
and at time t = 28 s, the car is again at the rear of the train.
Determine the speed of the car and the acceleration of the train.
Discussion: I haven't drawn a diagram, nor a velocity-time graph,
but it might be helpful to draw either or both if you're having
trouble following the solution to this problem.
I have constructed a table (below) to record the key information
given in the problem. It seems sensible to focus on the three
different times of interest, 0 s, 14 s, and 28 s.
There are two quantities that we need to determine, the speed
of the car and the acceleration of the train. We're not given a lot
of information, but the conditions that we are given involve the
positions of the car and train. This suggests that we use
kinematics equations involving displacement to relate the
quantities that we wish to calculate.
Solution: In the table below, variables labelled x represent the
positions of either the car or train at specific times.
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My strategy in solving this problem is to write down expressions for the
positions of the car at the times 14 s and 28 s, and then write down expressions
for the positions of the front of the train at the same times. Then I use the
conditions given in the problem to connect the positions of the car and the front
of the train.
Because the car moves at a constant speed,
Because the train's initial speed is zero, its displacement in the first 14 s is
The train's displacement from 14 s to 28 s is
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What is free fall?
Page 54, Problem 43 The greatest height reported for a
vertical jump into an air bag is 99.4 m by stuntman Dan Koko.
In 1948 he jumped from rest from the top of the Vegas World
Hotel and Casino. He struck the airbag at a speed of 39 m/s.
To assess the effects of air resistance, determine how fast he
would have been travelling on impact had air resistance been
absent.
Solution: I'll choose downward to be the positive direction.
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Solution: I'll choose downward to be the positive direction.
In taking the square root, we have to consider both signs.
We know the person falls down, in the positive direction,
so his impact velocity would have been +44.1 m/s if there
were no air resistance. The actual impact velocity is +39
m/s, so you can see that air resistance decreased the
impact velocity by about 5 m/s.
It's instructive to solve the problem again, but this time
choosing upward to be the positive direction. In this case,
both the acceleration due to gravity and the displacement
are negative, so the result will be
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In this case, the impact velocity is −44.1 m/s.
In both cases, the impact speed is 44.1 m/s. Thus, the
choice of positive direction does not affect the final
solution for the physically measurable quantity, the
impact speed.
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Page 54, Problem 44 A dynamite blast at a quarry launches a
chunk of rock straight upward, and 2.0 s later it is rising at a
speed of 15 m/s. Assuming air resistance does not exist, calculate
the speed of the rock (a) at launch, and (b) 5.0 s after launch.
Discussion: Choose "up" to be the positive direction. Then the
acceleration due to gravity is −9.8 m/s2, because the acceleration
due to gravity points down, which is the negative direction.
Let t = 0 represent the time at which the rock is launched. Thus,
we are looking for the speeds of the rock at t = 0 and t = 5 s.
Solution:
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Thus, the initial speed of the rock is 34.6 m/s and the speed of
the rock after 5 s is 14.4 m/s.
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Page 54, Problem 46 A ball is thrown vertically upward, and 8.0 s
later the ball returns to its point of release. Determine the ball's
initial speed.
Discussion: I'll choose "up" to be the positive direction again.
Solution 1: We are told that the displacement is zero after 8 s.
Substituting this information into the following equation will allow
us to calculate the unknown coefficient in the equation, the initial
speed.
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speed.
Thus, the initial speed is 39.2 m/s.
Solution 2:
Page 54, Problem 47 Two identical pellet guns are fired
simultaneously from the edge of a cliff. These guns impart an
initial speed of 30.0 m/s to each pellet. Gun A is fired straight
upward, with the pellet going up and then falling back down,
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upward, with the pellet going up and then falling back down,
eventually hitting the ground beneath the cliff. Gun B is fired
straight downward. In the absence of air resistance, how long
after pellet B hits the ground does pellet A hit the ground?
Discussion: I`ll choose upwards to be the positive direction, as
usual. Because there are two moving objects, we shall be careful
to distinguish the kinematical variables associated with each.
Let h represent the height of the cliff above the ground; thus, let
y = 0 represent the vertical position of the ground and let y = h
represent the vertical position of the top of the cliff.
Solution: We know the displacements of each pellet, we know
their initial velocities, we know their accelerations, and we wish
to determine information about the times at which they hit the
ground. This all suggests that we try working with the third
kinematics equation for each pellet:
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Page 54, Problem 49 A hot air balloon is rising upward
with a constant speed of 2.50 m/s. When the balloon is 3.0
m above the ground, the balloonist accidentally drops a
compass over the side of the balloon's basket. Determine
how much time elapses before the compass hits the
ground.
Solution: The key point to observe is that at the instant
that the compass is dropped, it shares the velocity of the
balloonist's hand, it rises upwards together with the
balloon and everything else in it, all at a speed of 2.50 m/s
in the upward direction. Thus, when the compass is
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in the upward direction. Thus, when the compass is
"dropped" it nevertheless continues to rise upwards for a
while, relative to the ground. Of course, relative to the
balloon, the compass does drop immediately, because the
balloon continues to rise upwards at a constant speed of
2.50 m/s, whereas the compass begins to slow down as
soon as it is released.
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We reject the negative value, as it is not relevant for our
problem. Thus, the compass reaches the ground 1.08 s after it is
released.
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Page 54, Problem 63 While standing on a bridge 15.0 m above
the ground, you drop a stone from rest. When the stone has
fallen 3.20 m, you throw a second stone straight down. What
initial speed must you give the second stone if they are both to
reach the ground at the same instant?
Discussion: I shall take upwards to be the positive direction. I
shall let y = 0 represent the position of the ground and y = 15 m
represent the position of the bridge.
Strategy: Let us calculate the time needed for the first stone to
fall 3.2 m, and then the time needed for it to fall from the bridge
right to the ground. The difference of these two times is the
time needed for the second stone to reach the ground, which
will help us determine its initial speed.
Remember that we assume no air resistance, so that the
acceleration of each stone is the acceleration due to gravity.
Solution:
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Page 54, Problem 64 A roof tile falls from rest from the top of a
building. An observer inside the building notices that it takes
0.20 s for the tile to pass her window, which has a height of 1.6
m. Determine the distance between the top of the window and
the roof.
Discussion: We assume that there is no air resistance, so that the
acceleration of the tile is the acceleration due to gravity. As
usual, we'll take upwards to be the positive direction.
Solution:
For the motion of the tile across the window,
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Now that we know the velocity of the tile at the top of the
window, we can calculate the initial position by focussing on
the motion above the window (i.e., from the initial position to
the top of the window):
How can we describe motion graphically?
Position-time graphs for zero acceleration (i.e., constant velocity)
Recall the key facts about the relations between position-time
graphs and velocity-time graphs:
• the slope of a position-time graph is the velocity
• the slope of a velocity-time graph is the acceleration
• the area under a velocity-time graph is the displacement
Here are some basic position-time graphs and velocity-time graphs:
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Velocity-time graphs for constant, non-zero acceleration
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Position-time graphs for constant, non-zero acceleration
For each of the seven position-time graphs displayed in the past
few pages, draw the corresponding velocity-time graph. Here's one
done for you to give you a sense for what I am asking you to do:
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Do you understand which of the four curved position-time
graphs given above matches with which of the four velocitytime graphs given above?
Note that we typically assume constant acceleration in this
course, which means that position-time graphs are either
straight lines or pieces of parabolas, and therefore velocitytime graphs are straight. If you know calculus, you can play
with position-time graphs that are more general.
Derivations of the kinematics equations for constant acceleration
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Derivations for calculus lovers ONLY:
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Derivations for equations (3) and (4) are as above.
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Page 56, Problem 86 A hot-air balloon is rising straight up at a
constant speed of 7.0 m/s. When the balloon is 12.0 m above the
ground, a gun fires a pellet straight up from ground level with an
initial speed of 30.0 m/s. Along the paths of the balloon and the
pellet, there are two places where each of them has the same
altitude at the same time. How far above the ground are these
places?
Discussion: As usual, I'll take the upwards direction to be
positive, and we'll pretend that there is no air resistance. I'll label
the ground as the zero-level for height.
There are several ways to solve this problem, as with virtually all
problems. I'll adopt a "functional" approach; I encourage all you
advanced students (including, and especially, physics majors) to
problems we think in terms of displacement, but in this problem I
want to shift the focus to thinking in terms of positions; in
particular, I want to highlight the fact that each moving object
has a position function. You can write the position of each
moving object as a function of time.
Because this problem asks specifically for positions, it makes
sense to think in terms of position functions.
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sense to think in terms of position functions.
So, write a position function for each moving object. The position
function of an object moving vertically has the form
Solution: Remember that the balloon moves at a constant
speed, and so its acceleration is zero. The pellet is a projectile,
in free fall, so its acceleration is the acceleration due to
gravity. Thus, the position functions of the moving objects are
and pellet are the same; to determine the corresponding
times, set the two position functions equal and solve for t.
Substituting these times into either of the position functions,
we obtain that the balloon and pellet have the same height
above the ground at the same time at the heights y = 16.185 m
and y = 40.672 m.
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Example: A hot-air balloon is rising straight up at a constant
speed of 7.0 m/s. When the balloon is 12.0 m above the ground
a coin is dropped from the balloon.
(a) Determine the maximum height above the ground reached
by the coin.
(b) Determine the time needed for the coin to reach the
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(b) Determine the time needed for the coin to reach the
maximum height.
(c.) Determine the time needed for the coin to reach the
ground.
Solution: Just as in the previous example, adopting a
"functional" approach makes for a smooth solution. That is,
think of the vertical position of the coin as a function of time,
and think of the vertical velocity of the coin as a function of
position.
First choose a positive direction; I'll choose "up" as positive, so
the acceleration due to gravity is negative, as it is in the
downwards direction.
One of the things people have trouble wrapping their minds
around at first is that when the coin is released its velocity is 7
m/s upwards. We called this the coin's "initial" velocity,
imagining that a stopwatch is started just as the coin is
released.
Of course, the coin immediately begins to gradually slow down
on its upward journey, because the acceleration opposes its
motion. The coin gradually slows down on its upward journey
until it stops for an instant at its peak height. Then the coin
begins to accelerate downward, picking up speed gradually and
continually as it falls.
In class we discussed that the key condition satisfied by the
coin's motion when it reaches its peak height is that its velocity
is zero then. We can use this condition to write an equation
that then allows us to solve parts (a) and (b).
Note, however, that the acceleration of the coin is never zero
while it is in the air. There is a common misconception that at
the peak of the coin's motion its acceleration is momentarily
zero. This is false. Our fundamental assumption is that
(ignoring air resistance, as we typically do) the acceleration of
the coin is constant while the coin is in the air, and therefore is
never zero while the coin is in the air. The coin's velocity is zero
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never zero while the coin is in the air. The coin's velocity is zero
at the peak, not the acceleration.
The condition for reaching maximum height is that
the velocity is zero.
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Note the symmetry of the graph of the position-time
function; the graph is a parabola, and the axis of
symmetry corresponds to the time at which the coin
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symmetry corresponds to the time at which the coin
reaches its peak height (which corresponds to the vertex
of the graph).
The intercepts of the position-time graph represent the
times when the coin is at the ground. The negative
solution of the equation is usually rejected, because only
the part of the position-time graph for positive times is
relevant for this coin's motion. Nevertheless, the
negative solution does have a meaning. What is it? (All
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Page 56, Problem 88 A football player, starting from rest at the
line of scrimmage, accelerates along a straight line for a time of
1.5 s. Then, during a negligible amount of time, he changes the
magnitude of his acceleration to 1.1 m/s2. With this acceleration,
he continues in the same direction for another 1.2 s, until he
reaches a speed of 3.4 m/s. Determine the value of his
acceleration (assumed to be constant) during the initial 1.5-s
period.
this one. My result is 1.39 m/s2.
Page 56, Problem 81 A woman and her dog are out for a
morning run to the river, which is 4.0 km away. The woman runs
at 2.5 m/s in a straight line. Her dog is unleashed and runs back
and forth at 4.5 m/s between his owner and the river, until the
woman reaches the river. Determine the distance travelled by
the dog.
Solution: This is a neat little problem. The dog runs in a
complicated back-and-forth motion, and if we assume that the
dog has zero width, and makes turns instantly, then there are an
infinite number of legs to the dog's journey. ("Balto, a dog
cannot make this journey alone; but … maybe, a wolf of zero
width can!")
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width can!")
For those of you who will take MATH 1P02 or 1P06 next
semester, you'll learn about infinite series, so you'll have a shot
at solving the problem using an infinite series, but there has to
be a faster and more way to solve the problem … see what you
can do with this one.
The problem is reminiscent of a famous story involving
mathematician John von Neumann, who was famous for being
able to do complex calculations mentally. At a party, another
mathematician gave him a similar problem; von Neumann
thought for a few seconds, and then gave the correct answer.
"Most people try to sum the infinite series," commented the
other mathematician, assuming that because von Neumann
answered so quickly he must have used the quick, insightful
method. Nonplussed, von Neumann replied, "Yes, that's what I
did." Yes, he was a very smart guy.
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