Motion Concepts Explained

2
Motion in One Dimension
CLICKER QUESTIONS
Question A2.01
Description: Introducing the concept of average speed.
Question
Michael is going to the store 6 miles away. He rides his bike at 12 mph for the fi rst half of the trip, then
walks at 4 mph for the remainder.
Michael’s average speed for the trip to the store is closest to:
1. 2 mph
2. 4 mph
3. 6 mph
4. 8 mph
5. 10 mph
6. 12 mph
7. 14 mph
8. Exactly halfway between two of the values above
9. Impossible to determine
Commentary
Purpose: To introduce or hone the concept of average speed.
Discussion: The average speed is the total distance traveled divided by the total time needed to travel that
distance.
In this case, we know the total distance is 6 miles, but we do not know the total time.
If you assume that “fi rst half of the trip” means after 3 miles, then it takes Michael 1/4 hour to go the fi rst
3 miles (at 12 mph) and 3/4 hour to go the second 3 miles (at 4 mph), for a total of 1 hour and an average
speed of 6 mph.
If you assume that “fi rst half of the trip” means that he spent the same amount of time at each speed, then
the average speed is the average of 12 mph and 4 mph, or 8 mph. Let’s get this another way: Let t be the
time needed to complete each half of the trip (in hours). The total distance is therefore (12t + 4t), which
must be equal to 6 miles, or 16t = 6. But the average speed is simply the total distance divided by the total
time, or 6/(2t) = 3/t = 8 (mph).
Since the meaning of “the fi rst half of the trip” is ambiguous here—does it refer to distance or time?—
answers (3), (4), and (9) are all defensible.
17

18 Chapter 2
Key Points:
• The average speed for a journey is equal to the total distance traveled divided by the total time
required.
• Sometimes everyday language like “the fi rst half of the trip” is not precise enough for physics use.
For Instructors Only
Most students will simply average 12 mph and 4 mph to get 8 mph, so even though there is a valid
assumption that will lead to this answer, most students will not have thought much about it. You will need
to fi nd out why students picked each choice.
Note that the “average speed” is properly interpreted as a “time-weighted” average, though most students
will not appreciate what this means. On a graph of speed vs. time, the average is the area below the graph
(total distance) divided by the length of the base (total time).
Some students might have learned a technique for correctly computing average speed without having any
clue why it works. They break the trip into well-chosen equal time intervals, then average the (constant)
speeds during the intervals. In the case of equal distances, if we break the trip into four 15-minute intervals,
the speeds are 12, 4, 4, and 4 mph, for an average of (12 + 4 + 4 + 4)4 = 6 (mph).
Question A2.02
Description: Honing the concept of average speed.
Question
Wendy walks 10 m in one direction at 2 m/s, then runs 6 m in the same direction at 6 m/s. Next, she stops
for 4 seconds, and fi nally walks in the opposite direction at 4 m/s for 6 seconds.
Wendy’s average speed is closest to:
1. 1
2 m s
2. 1 m s
3. 1 1
2 m s
4. 2 m s
5. 2 1
2 m s
6. 3 m s
7. 3 1
2 m s
8. 4 m s
9. the negative of one of the choices above
10. impossible to determine
Commentary
Purpose: To hone the concept of “average speed.”

Motion in One Dimension 19
Discussion: The “average speed” of someone is the total distance traveled divided by the total time needed
to travel that distance.
In this case, Wendy walks 10 m, then runs 6 m, stops briefl y, and fi nally walks 24 m, for a total distance
traveled of 40 m. Direction does not matter. It takes Wendy a total of 16 seconds to travel that distance, for
an average speed of 2.5 m/s.
Note that you cannot ignore the 4 seconds during which Wendy is not moving. That is still part of her
motion.
Key Points:
• The average speed of an object is the total distance it travels (the total length of the path along which it
moves, regardless of direction) divided by the total time of travel.
• Average speed is never negative.
• Average speed is not the magnitude of “average velocity.”
Answer (1) is the magnitude of the average velocity.
Answer (2) is the average of the four velocities mentioned (2, 6, 0, and −4).
Answer (3) is closest to the average of the three nonzero velocities (2, 6, and −4).
Answer (6) is the average of the four individual speeds (2, 6, 0, and 4).
Answer (7) is closest to the average speed if the four seconds spent at rest are ignored. (Some people ignore
“rest stops” on long car trips when computing average speed.)
Answer (8) is the average of the three nonzero speeds (2, 6, and 4).
Answer (9) would fi t the average velocity, − 1
2 m s.
Answer (10), “impossible to determine,” might be chosen because directions are not given (which is
irrelevant) or because the travel segments are described inconsistently (which makes the question more
diffi cult but not impossible).
Of course, students may make choices based on algebra mistakes as well.
Some alternate approaches to answering the question:
• Sketch speed vs. time, or otherwise break the process into sixteen 1-second intervals. Average these
16 speeds to get 2 1
2 m s .
• Break the 16 seconds into two 8-second intervals. Wendy travels 16 m for an average speed of 2 m/s
during the fi rst and 24 m for an average speed of 3 m/s during the second. The average of these is
2 1
2 m s .
• Break the process into any number of equal time intervals and determine the average speed during
each. Since average speed is a “time average,” you can average these averages.
A sketch of speed vs. time (or even velocity vs. time, suitably analyzed) should help students organize
information and compute the correct value.

20 Chapter 2
Some students may question whether someone can go immediately from one speed or velocity to another.
The quick answer is that we are ignoring these short periods of acceleration. A longer answer is that the
given speeds are averages for the distances or times given. So, for example, during the second leg of
the process, Wendy would actually reach a speed larger than 6 m s during the 6 m she runs. The bottom line
is that she completes the distance in 1 second.
Additional Discussion Questions
1. What is Wendy’s average velocity?
2. Sketch Wendy’s velocity vs. time.
Question A2.03a
Description: Distinguishing speed and velocity, and considering the speed of a vertical projectile at the
top of its trajectory.
Question
A ball is thrown straight up into the air. Its position at 7 instants of time are shown in the fi gure;
the maximum height is reached at position 4. At which of the labeled points is the speed of the
ball smallest?
1. point 1
2. point 2
3. point 3
4. point 4
5. point 5
6. point 6
7. point 7
8. Exactly 2 of the points shown
9. More than 2 of the points shown
Commentary
4
3
2
Purpose: To probe your understanding of the difference between speed and velocity, and to establish the
idea that the velocity (and speed) of a vertically launched projectile at the top of its trajectory is zero for
one instantaneous point in time.
Discussion: An object’s velocity is the rate of change of rate of change of its position and describes
both how fast the object is moving (“speed”) and in which direction (positive or negative sign for one
dimension, vector direction for two or three). Speed is the magnitude of velocity (absolute value in
one dimension), and is never negative.
As the ball moves upward, it slows down under the infl uence of gravity; that means the velocity (which
is positive) gets smaller and smaller. When it reaches zero, the ball is at rest for one instant in time, and
immediately starts moving downward with a negative velocity. As the ball falls faster and faster downward,
the velocity gets more negative (larger speed in the negative direction). Thus, the speed is positive every-where
except for point (4) at the very top, where it is zero instantaneously.
If the question had asked where the velocity is least, the correct answer would be (7): the most negative
number is the least.
5
6
1 7

Key Points:
• Velocity is the rate at which position is changing, and has a magnitude and a direction.
• Speed is the magnitude of the velocity, and is never negative.
• An object launched vertically upward has a positive velocity while moving upward (assuming we
defi ne our coordinate system so up is the positive direction), negative velocity while falling back down,
and zero velocity for just an instant as the velocity changes from positive to negative.
This question exists largely to set up the next two questions in this three-question set.
Use this question to ascertain whether your students understand the difference between velocity vs. speed,
and that both are momentarily zero at the top of the trajectory. Then move on to the next to questions,
which are likely to be much more contentious and productive.
One issue that may need discussion is what “smaller/larger” and “less/greater” mean in the context of a
number line with positive and negative numbers. Students may not be aware that “smaller” means “closer
to zero” or “smaller in magnitude,” while “less” means “closer to the negative end of the number line.”
Question A2.03b
Description: Honing the concept of acceleration for a vertical projectile, and probing for a common
misconception.
Question
A ball is thrown straight up into the air. Its position at 7 instants of time are shown in the fi gure;
the maximum height is reached at position 4. At which of the labeled points is the acceleration
of the ball smallest?
1. point 1
2. point 2
3. point 3
4. point 4
5. point 5
6. point 6
7. point 7
Commentary
3
2
Purpose: To explore your understanding of the concept of acceleration, probe for the common miscon-ception
that an object’s acceleration is zero whenever its velocity is momentarily zero, and develop your
understanding of “free fall.”
4
5
6
1 7

22 Chapter 2
Discussion: While traveling up and back down, the only signifi cant force affecting the ball is gravity. We
call this situation “free fall” (even for the upward part of the motion). The gravitational force is a constant,
depending only on the mass of the ball and not on the ball’s position or motion. According to Newton’s
Second Law (F
net = ma), the acceleration of a body is proportional to the net force acting upon it. Since the
force on a body in free-fall is constant, so is the acceleration. Thus, answer (9) is best.
(You’ll be learning more about Newton’s Second Law soon; for now, all you need is the basic ideas that
acceleration is proportional to total force, the only force acting is gravity, and gravity is constant.)
You might be tempted to think that the acceleration is zero at the very top of the trajectory, just like the
velocity. However, if the acceleration were zero, that would mean the velocity is not changing; so if the
velocity were zero, it would remain zero, and the ball would just hover there. For a ball resting on a table,
the acceleration and velocity are both zero, which is why the ball doesn’t move.
You might argue that the acceleration is not constant, but is larger at point 1 than point 7. This may be
correct, if your argument is based on aerodynamic drag (air resistance): on the way up, drag opposes the
motion and thus exerts a downward force that augments gravity; on the way down, drag opposes the motion
and thus exerts an upward force that opposes gravity. Furthermore, that force is stronger when the ball is
moving faster, at the start and end of the trajectory. So, answer (7) is defensible.
If you don’t make the approximation of “local gravity” but instead take into account the fact that the gravi-tational
force on the ball gets weaker as the ball gets farther away from the Earth, then you could argue that
the acceleration is smallest at point (4), the top of the trajectory. However, this effect is incredibly small, far
smaller than air resistance. If you’re going to be this exacting, the effect of air resistance will dominate, and
(7) is a better answer than (4).
Key Points:
• An object in “free fall” (i.e., traveling under the infl uence of gravity alone) has a constant downward
acceleration.
• If an object has zero velocity and zero acceleration at the same time, it is remaining stationary.
• The best answer to this question (and many others) depends on what approximations you make.
Learning what approximations physicists typically make, and when they make them, is an important
part of learning physics.
Worried about a problem that requires referring to Newton’s Second Law before actually presenting it?
Remember, these questions are to stimulate and organize learning, not to “test” students on material already
covered. “Fair” is irrelevant; “productive” is the goal.
Don’t be dismayed if students don’t neglect air resistance. Part of learning physics is learning to make the
standard assumptions and approximations that practicing physicists do. Explicitly discussing such assump-tions
and considering how making or not making them affects answers helps them to do so. If we assert that
the acceleration is the same everywhere and we don’t explicitly point out that we’re ignoring air resistance
and that what we said is only true in that approximation, we can confuse students rather than helping them.
Of course, including air resistance isn’t the only reason why a student might choose answer (7). As always,
our fi rst task when we see the answer histogram is to elicit as complete as possible a spectrum of students’
arguments for their answers.

Question A2.03c
Description: Distinguishing, relating, and reasoning with kinematic quantities.
Question
A ball is thrown straight up into the air. Its position at 7 instants of time are shown below; the
maximum height is reached at position 4. At which of the labeled points is the speed of the ball
largest?
1. point 1
2. point 2
3. point 3
4. point 4
5. point 5
6. point 6
7. point 7
Commentary
4
3
2
1 7
Purpose: To enrich your understanding of the relationship between acceleration, velocity, displacement,
and position; develop your qualitative reasoning skills; and demonstrate the power of reasoning with
graphs.
5
6
Discussion: The ball slows down as it rises, comes momentarily to rest, and then falls back down with
increasing speed. The ball’s speed must therefore be largest at either point 1, point 7, or both. Neglecting
air resistance, it’s easy to show that the speed of the ball is the same at points 1 and 7 using the principle of
Conservation of Energy. However, we won’t be seeing that until later in the course. How can we convince
ourselves that the speed is the same at these two points using the physics we already know?
In the previous problem (22b), we established that the ball’s acceleration is constant since the only force
acting on it is gravity, which is constant. Acceleration is the rate of change of velocity, which means that
acceleration is the slope of a velocity vs time graph. If acceleration is constant, the velocity vs. time graph
must be a straight line.
vy
0
up
down t
Velocity is the rate of change of position, which means that the area under the velocity vs. time graph indi-cates
the displacement of the object: the change in its position (xf − xi). Area above the t-axis indicates posi-tive
displacement (increase of position), area below indicates negative displacement (decrease of position).
If the ball is to return to the point at which it started, its total displacement must be zero. This means that
the top-left triangle must have the same area as the bottom-right triangle. The only way for this to happen
is if the triangles are the same size, which means that the velocity at the end has the same magnitude as the
velocity at the beginning. Thus, the speed at points 1 and 7 must be the same.

24 Chapter 2
If we do not neglect air resistance, acceleration isn’t quite constant, so the velocity vs. time graph is slightly
curved, and the fi nal speed won’t be quite the same as the initial speed to make the total displacement be zero.
Key Points:
• In the absence of air resistance, a vertical projectile lands with the same speed at which it was
launched.
• Acceleration is the slope of a velocity vs. time graph, and displacement (change in position) is the area
under it.
• Graphs are helpful tools for reasoning about situations and answering questions. Understanding how to
interpret the slope of and area under a graph is powerful.
Though we might be tempted to simply assert that the ball’s speed must be the same at start and fi nish—
perhaps considering it almost self-evident—it’s not necessarily so obvious to students, especially with the
tools they currently have to work with.
If we want students to use graphs, free body diagrams, and other nonalgebraic representations, we must
give them problems where these approaches are clearly superior, and also model their use.
Question A2.04a
Description: Understanding the sign of velocity and acceleration.
Question
A ball is rolled up an incline so that it goes part-way up and then rolls back down. Which of the graphs below
could represent its acceleration vs. time from the instant it is released until it returns to where it started?
1.
2.
3.

4.
5.
6.
7.
8.
9. None of the graphs
10. Two or more of the graphs
Commentary
Purpose: To hone the concept of acceleration, focusing on its vector nature, and how to represent it
graphically. In particular, this question targets the common misconception that positive acceleration means
“speeding up” and negative acceleration means “slowing down.”
Discussion: A ball rolled up an incline slows down, stops at the top, then speeds up again as it rolls back
down. It is common but incorrect to think that the acceleration is negative while the ball slows down and
positive while the ball speeds up.
Acceleration is defi ned as the change in velocity. As the ball rolls up the incline, its velocity points in its
direction of motion, parallel to the plane and uphill. As the ball slows down, the velocity vector gets shorter.
The change in the velocity vector between two subsequent times is therefore a vector pointing parallel to
the plane and downhill. So the acceleration points downhill.

26 Chapter 2
As the ball rolls down the hill, its velocity again points in the direction of motion, which is now downhill.
It is speeding up, so the velocity vector is getting longer. The change in the velocity vector between two
subsequent times is therefore a vector pointing parallel to the plane and downhill. So the acceleration again
points downhill.
Even at the very top of its motion, when the ball stops rolling up and starts rolling back down, its velocity
is changing from a vector pointing up the plane to one pointing down the plane. Here too the acceleration
points downhill. Throughout its entire motion, the ball’s acceleration is nonzero and points down the plane.
Therefore, answers (1), (2), (4), (5), (6), (7), and (8) cannot be valid.
Answer (9) can be valid, if we choose a coordinate system so that the positive direction is down the plane.
This may be an unusual choice, but it is valid. Keep in mind that coordinate systems are arbitrary math-ematical
constructs we defi ne to help us solve problems; we can orient them however we wish. Since the
question asks which of the graphs could represent the ball’s acceleration vs. time, answer (9) is the best
choice.
Key Points:
• Acceleration is the rate of change of velocity.
• If an object is traveling in a straight line and slowing down, acceleration points in the opposite
direction of its motion. If it is speeding up, acceleration points in the direction of motion.
• “Positive” and “negative” accelerations refer to the direction relative to a coordinate system, not to
speeding up or slowing down.
• Coordinate systems are arbitrary and may be oriented however we wish, though some choices are more
common and convenient than others.
This is fi rst of two similar questions exploring graphs of velocity and acceleration. We recommend present-ing
both questions back-to-back, collecting answers for each, before discussing or revealing anything about
either. (Question A3.07 is similar, and is intended for use when introducing curvilinear motion. It explores
the concept of tangential acceleration, which is close to many students’ intuitive idea of acceleration.)
Most students generally choose answer (2), since they associate “slowing down” with “negative
acceleration” and “speeding up” with “positive acceleration.” The instructor must lead students to explicitly
articulate this idea, and then convince them that it is inconsistent with the defi nition of acceleration.
Some students, especially those who are trying to deal with coordinate frames and just slightly missing the
mark, will pick (1), or perhaps both (1) and (2). These students may be more sensitive than most to the fact
that the “positive” direction is positive. For them, choosing down to be positive makes (1) valid.
Answer (9) is a likely choice for students who understand that the acceleration will not change magnitude
or direction, but assume that the positive direction must be up the plane. Others, less confi dent in their
thinking, will assume that they are wrong and pick one of the other answers.
Graph (4) is a valid graph of speed vs. time, and (5) is a valid graph of velocity vs. time. If you are
discussing the two questions in the set together, this is a good connection to make. If you have not
presented 23b yet, we recommend not emphasizing this point until you have.
Graph (6) is the negative of the speed. (7) is position vs. time with the initial height chosen as the origin,
and (8) is position vs. time with the topmost point chosen as the origin.

Question A2.04b
Description: Understanding the sign of velocity and acceleration.
Question
A ball is rolled up an incline so that it goes part-way up and then rolls back down. Which of the graphs
below could represent its velocity vs. time from the instant it is released until it returns to where it started?
1.
2.
3.
4.
5.
6.

28 Chapter 2
7.
8.
9. None of the graphs
10. Two or more of the graphs
Commentary
Purpose: To hone the concept of velocity, focusing on its vector nature and how to represent it graphically.
Discussion: We expect the ball to slow down, change direction, then speed up in the opposite direction, all
with constant acceleration. There are two valid ways of representing this motion, though only one of them
is shown. Graph (5) shows an object moving in the negative direction but slowing down, then stopping
(where the line crosses the t-axis) and speeding up in the positive direction. If we choose a coordinate sys-tem
such that downhill is positive, and uphill is negative, this is a possible graph for the motion of the ball.
So (5) is a defensible answer.
If uphill were chosen as the positive direction, the velocity vs. time graph would be a straight line with
a negative slope, starting above the t-axis and ending below it. This choice is not included among the
answers.
Graph (1) depicts a ball that travels up the incline with constant velocity, and then suddenly and instantane-ously
reverses. This graph would be more appropriate for a ball that rolls along a horizontal fl oor and then
rebounds off of a wall. (2) is the same, but with positive and negative directions reversed, as if downhill
were chosen as the positive direction.
Graph (4) is a valid representation of speed vs. time, but not of velocity vs. time. Since the function
depicted is always positive except for the infl ection point, the object would always be moving in the same
direction. This graph might represent a car that slows down, stops momentarily at an intersection, and then
speeds up again in the same direction as before.
Graph (6) cannot be speed vs. time for any coordinate system: speed is never negative. As with (4), it could
represent a car slowing and stopping momentarily at an intersection and the proceeding in the same direc-tion,
if the car were traveling in the negative direction of our coordinate system for the entire time.
Note that acceleration vs. time is the slope of velocity vs. time. For graph (5), the slope is constant and
positive at all times, so the corresponding graph of acceleration vs. time must be a constant, positive value,
such as graph (3).
Key Points:
• The sign of an object’s velocity indicates whether it is moving in the “positive” or “negative” direction
as defi ned by a chosen coordinate system. The magnitude indicates the object’s speed.

• If an object experiences constant acceleration, its velocity vs. time graph must be a straight line whose
slope equals the value of acceleration.
• There is no one “right” coordinate system for a situation or problem. For convenience and out of habit,
we generally choose the positive direction to be upward or to the right, but that is not necessary.
This is fi rst of two similar questions exploring graphs of velocity and acceleration. We recommend present-ing
both questions back-to-back, collecting answers for each, before discussing or revealing anything about
either. (Question A3.07 is similar, and is intended for use when introducing curvilinear motion. It explores
the concept of tangential acceleration, which is close to many students’ intuitive idea of acceleration.)
Most likely, the majority of your students will select answer (9): “None of the graphs.” They are most likely
looking for a graph that begins with a positive value of velocity sloping down to a negative value at the
end, because they are using the convention that “up” is positive. A primary objective of this question is to
help them appreciate that this is merely a choice, not a necessity. If “down” is chosen to be positive, then
(5) is a valid choice, and the question asks which of the graphs could represent the ball’s velocity vs. time.
Students should not be expected to know that the ball will slow down going up and speed up rolling back
down, rather than instantaneously reversing direction. Changes in speed are diffi cult to detect in real-life
observation, and students who have never taken physics before may not realize the ball is in fact slowing
down as it rolls up the incline. If any students choose answers (1) or (2) for this reason, it is important
to confi rm that they are correctly representing their physical model of the situation graphically, though
their physical intuition needs refi nement. Simply saying those answers are wrong risks confusing such
students—they might think they were wrong for the wrong reason.
Question A2.05a
Description: Honing the concept of position.
Question
Ralph walks 3 m to the left in 2 seconds, then 4 m to the right in 3 seconds. Next, he stops for 3 seconds,
and fi nally walks 5 m to the left in 4 seconds.
Ralph’s fi nal position is closest to:
1. 0 m
2. 2 m
3. 4 m
4. 6 m
5. 8 m
6. 10 m
7. 12 m
8. 14 m
9. The negative of one of the choices above

30 Chapter 2
Commentary
Purpose: To hone your understanding of “position.”
Discussion: An object’s “position” is its location relative to a coordinate system. We know a lot about the
motion of Ralph, but we do not know his position, because we do not know where to place the origin of the
coordinate system. In other words, we do not know Ralph’s initial position. It is tempting, but unjustifi ed,
to assume that he begins at the origin.
Also, the question does not defi ne which direction is “positive.” Even if the question told us that Ralph
began at the origin, we would not know whether he ended up at 4 m or −4 m. “Positive to the right” is a
convention we use when drawing graphs on paper, but does not necessarily describe the coordinate system
of Ralph’s world.
Key Points:
• An object’s “position” is its location relative to an origin.
• The origin of a coordinate system is not necessarily an object’s starting point.
• Knowing changes in position is not enough to determine fi nal position unless we know the initial
position as well.
This is the fi rst of three related questions that help students distinguish and relate position, distance
traveled, and displacement.
Students choosing answer (3) are most likely assuming Ralph begins at the origin. Students choosing
answer (9) may be doing the same, and assuming that “left” means “in the negative direction.”
To help students understand why an object’s starting position might not always be at the origin, ask how
they would describe a situation with two individuals beginning in different places.
Question A2.05b
Description: Honing the concept of distance traveled.
Question
The distance Ralph travels is closest to:
1. 0 m
2. 2 m
3. 4 m
4. 6 m
5. 8 m
6. 10 m
7. 12 m
8. 14 m

Commentary
Purpose: To hone your understanding of “distance traveled.”
Discussion: The “distance traveled” by someone is not how far the person ends up from his original loca-tion,
but the total amount of distance covered during the process of moving. If you walk from home to the
store and then back again, your “distance traveled” is twice the distance to the store, even though you end
up where you started. When driving a car, the distance traveled is the change in odometer reading.
We often call this the “total” distance traveled, just to be clear.
In this case, Ralph walks 3 m, then 4 m, and fi nally 5 m, for a total of 12 m. Direction does not matter, nor
does his starting position.
Key Points:
• The “distance traveled” is the total distance moved during a specifi ed process. It is the sum of the
distances traveled during each leg of a trip.
• The distance traveled is always positive. Direction does not matter.
• The distance traveled depends on the path taken between two points, but it does not depend on how
long it takes to complete.
This is the second of three related questions that help students distinguish and relate position, distance
Students who choose answer (3) are likely determining the displacement, or its magnitude, rather than the
distance traveled. They may think of this as the “net” distance traveled.
Students choosing answer (9) may be determining the displacement and assuming positive is to the right, or
applying some kind of sign convention to the distance, or perhaps a different error.
Question A2.05c
Description: Honing the concept of displacement.
Question
Ralph’s displacement is closest to:
1. 0 m
2. 2 m
3. 4 m
4. 6 m
5. 8 m
6. 10 m
7. 12 m
8. 14 m

32 Chapter 2
Commentary
Purpose: To hone your understanding of “displacement.”
Discussion: An object’s displacement is the change in its position: fi nal position minus initial position. It has
direction associated with it, and does not depend on the path taken between the initial and fi nal points or on
the time interval. Mathematically, if x describes an object’s position, we represent its displacement as Δx.
Ralph ends up 4 m to the left of his original position, so his displacement is “4 m, left.” If the positive
direction is chosen to be to the left, then “4 m” (answer 3) is acceptable; if positive is chosen to be to the
right, then “−4 m” (answer 9) is acceptable.
Key Points:
• An object’s “displacement” is the change in its position.
• Displacement has a direction associated with it, represented by the sign of the displacement (positive or
negative) in one dimension and by a vector direction in two or more dimensions.
• Displacement does not depend on how the object gets from the initial to the fi nal point.
This is the third of three related questions that help students distinguish and relate position, distance
Students might not like that there are two acceptable answers, and that the correctness depends on the
assumption made about the positive direction.
Question A2.06a
Description: Honing the concept displacement, and linking to graphical representations.
Question
Amy, Brad, and Cate are walking (or running) along a straight line as represented below.
Which people have the same displacement?
1. None; they all have different displacements.
2. Amy and Brad
3. Amy and Cate
4. Brad and Cate
5. All three are the same.
Commentary
Purpose: To hone your understanding of displacement.
Amy
Brad
t
Cate
2
1
0
0 1 2 3 4 5
time (s)
position (m)
6
1
2
x

Discussion: Displacement is the change in position. The intermediate path does not matter, but the overall
direction does.
Amy starts at x = −1 m and ends at x = +2 m, for a displacement of +3 m. Brad starts at x = −2 m and ends
at x = +1 m, for a displacement of +3 m. Cate starts at x = +1 m and ends at x = −2 m, for a displacement of
−3 m.
So, even though Amy and Brad have completely different paths, and also start and end at different points,
each person’s position changes by 3 m in the “positive” direction, so they have the same displacement.
Key Points:
• Displacement Δx is the change in position, Δx = x − x final initial.
• Displacement has a magnitude and a direction. In one dimension, a positive or negative sign indicates
direction. In two or more dimensions, we use a vector.
• Displacement does not depend upon the details of the motion between two points, or where they are
relative to the origin.
This is one of six questions about this graph. You do not need to use them all, or in any particular order.
Students who choose answer (5) are probably considering only the magnitude of the displacement.
Students who choose answer (1) might be confusing displacement with distance traveled or with fi nal posi-tion,
or otherwise taking the path traveled into account.
(Brad and Cate both have the same distance traveled—3 m—though in opposite directions. Amy travels a
longer total distance for the same magnitude of displacement.)
Depending on how much experience students have working with graphical representations, students may
have diffi culty interpreting these position vs. time graphs. (The more trouble they have, the more they need
to wrestle with this!) They might think they depict actual two-dimensional paths ( y vs. x) as seen from
above. Students who think this might still choose Amy and Brad as having the same displacement. A ques-tion
specifi cally asking students to compute the displacement or the distance traveled will help tease this
apart.
Question A2.06b
Description: Honing the concept of displacement, and linking to graphical representations.
Question
Amy, Brad, and Cate are walking (or running) along a
straight line as shown below.
Which people have the largest displacement?
1. only Amy
2. only Brad
3. only Cate
Amy
Brad
t
Cate
2
1
0
0 1 2 3 4 5
time (s)
position (m)
6
1
2
x

34 Chapter 2
4. Amy and Brad
5. Amy and Cate
6. Brad and Cate
7. All are the same
Commentary
Purpose: To hone your understanding of displacement.
Discussion: Displacement is the change in position. The intermediate path does not matter, but the overall
direction does.
Amy starts at x = −1 m and ends at x = +2 m, for a displacement of +3 m. Brad starts at x = −2 m and
ends at x = +1 m, for a displacement of +3 m. Cate starts at x = +1 m and ends at x = −2 m, for a
displacement of −3 m.
Thus, assuming that “largest” refers to the magnitude of the displacement, all three have the same magni-tude
of 3 m.
Key Points:
• Displacement Δx is the change in position, Δx = x x final initial − .
• Displacement does not depend upon the details of the motion between two points, or where they are
relative to the origin.
• The “largest” of a set of numbers usually refers to the one with the greatest magnitude, while the
“greatest” usually refers to the most positive or least negative (i.e., the one farthest towards the positive
end of the number line). People are not always consistent about this, however.
Amy and Cate end up farthest from the origin, each 2 m away, with Amy at a positive position and Cate at
a negative position. Students who choose answers (1) or (5) could be confusing displacement with position
or magnitude of position.
Amy and Brad have the same positive displacement. Students who choose answer (4) are perhaps thinking
that “+3” is larger than “−3.” (We typically interpret “largest,” “smallest,” “larger than,” and “smaller than”
as referring to magnitude and “greatest,” “least,” “greater than,” and “less than” as referring to relative
position along a number line. So, for example, −3 is “less than” but also “larger than” +1.)
Amy walks the farthest distance. Students who choose answer (1) might think that they are being asked for
the distance traveled.
Students unaccustomed to position−time graphs may interpret the graph lines as two-dimensional paths or
y vs. x trajectories. With this interpretation, all three people still have the same magnitudes of displacement.
You can determine whether this mistake is occurring by asking students during post-question discussion
what numerical value of the displacement they found. If they say that the displacements are all about 7 m
long, then they are probably interpreting these graphs as y vs. x instead of x vs. t.

Question A2.06c
Description: Honing the concept of distance traveled, and linking to graphical representations.
Question
Amy, Brad, and Cate are walking (or running) along a straight line as shown below.
Amy’s distance traveled is closest to:
1. 1 m
2. 3 m
3. 5 m
4. 7 m
5. 9 m
6. 11 m
7. 13 m
8. 15 m
9. 17 m
10. Exactly halfway between two of the values above
Commentary
position (m)
Purpose: To hone your understanding of “distance traveled.”
Amy
Brad
t
Cate
2
1
0
x
0 1 2 3 4 5
time (s)
6
1
2
Discussion: The distance traveled by an object is the total length of the path followed. It is a non negative
number, with no direction associated with it. The distance a car travels is the increase in its odometer
reading.
Amy starts at x = −1 m and fi rst walks to x = +2 m, a distance of 3 m. Then she walks to x = +1 m, a
distance of 1 m. Next, she stands still at x = +1 m for 2 seconds. Finally, she walks back to x = +2 m,
a distance of 1 m. Thus, the total distance traveled is 3+1+0+1 = 5 m.
Note that these graphs are not y vs. x, but position x vs. time t. These people are walking along a single
straight line, and the coordinate along this straight line is x. This representation is useful and common, so
you should make sure you understand it.
Brad and Cate travel each travel a total distance of 3 m.
Key Points:
• “Distance traveled” refers to the total length of the path followed by an object. It is a magnitude with
no sign or direction.
Students who choose answer (5) or perhaps (6) may be interpreting the graphs as y vs. x (top views). If
the graph did show y vs. x, Amy’s distance traveled would be about 9.8 m (and her displacement is about
6.7 m).

36 Chapter 2
Question A2.06d
Description: Honing the concept of average speed, and linking to graphical representations.
Question
Which people have the same average speed during the
time period shown?
1. None; they all have different average speeds.
2. Amy and Brad
3. Amy and Cate
4. Brad and Cate
5. All three are the same.
Commentary
Purpose: To hone your understanding of average speed.
t
Discussion: The average speed is the total distance traveled divided by the total time needed to travel that
distance. Direction does not matter.
The time period is the same for all three people, so we can focus on total distance traveled. If two people
travel the same distance, they must have the same average speed.
Amy travels 5 m in 6 seconds. Brad and Cate travel 3 m in 6 seconds, even though it is in opposite direc-tions
and with completely different pattern of speeds. Thus, Brad and Cate have the same average speed.
Key Points:
• An object’s average speed is the total distance it travels divided by the total time it is traveling.
• Average speed is not the magnitude of average velocity!
Students who answer (1) may be interpreting these graphs as y vs. x, since with that interpretation all three
have different path lengths.
Students who answer (5) may be fi nding the magnitude of the average velocity, since that is the same for all
three people (0.5 m/s).
Students are likely to compute numerical values for the average speeds, not realizing that since the times
are the same they can focus only on distance traveled. We recommend asking about this during discussion,
so they may realize they could have saved themselves some work by thinking more.
If students are having diffi culty with this, sketching speed vs. time plots may help.
Amy
Brad
Cate
2
1
0
0 1 2 3 4 5
time (s)
position (m)
6
1
2
x

An alternative technique students may have learned for computing average speed is to break the time
period into six 1-second intervals, and fi nd the speed during each. Then, they can average those six values.
(We don’t recommend this approach, as it obscures the concepts underlying average speed and doesn’t
generalize easily to situations with nonconstant velocity.)
Question A2.06e
Description: Honing the concept of instantaneous velocity, and linking to graphical representations.
Question
Which people are moving toward the origin at t = 2 s?
1. None
2. Amy only
3. Brad only
4. Cate only
5. Amy and Brad
6. Amy and Cate
7. Brad and Cate
8. All three
Commentary
2
1
0
x
0 1 2 3 4 5
time (s)
position (m)
1
2
Purpose: To hone the understanding of instantaneous velocity, in particular of its direction.
Amy
Brad
t
Cate
6
Discussion: We often use “positive” or “negative” to describe the direction of the velocity, but “positive”
does not always mean “away from the origin” and “negative” does not always mean “toward the origin.”
These are true only when the position is positive. In this case, Amy and Cate have positive positions and
negative velocities at t = 2 s, so they are moving toward the origin, i.e., moving toward x = 0. Cate reaches
the origin at t = 3 s.
But Brad is also moving toward x = 0. His position is negative and his velocity is positive. He reaches the
origin at t = 5 s.
Note that Amy does not return to the origin after being there earlier, but this does not mean she is not mov-ing
toward the origin at t = 2 s. If she had continued to walk at −1 2 m s , she would have reached the
origin at t = 5 s, but instead she stops at t = 3 s and then starts to move away from the origin at t = 5 s.
Key Points:
• For motion in one dimension, a “positive” velocity is away from the origin if the position is positive
and towards the origin if the position is negative. (Vice-versa for a “negative” velocity.)
Answer (6) can be surprisingly common: students often assume that people are on the positive side of an
origin, even if it might seem obvious to you that some are and some aren’t.

38 Chapter 2
Students who choose answer (7) or (4) may be thinking that Amy isn’t moving towards the origin because
she doesn’t actually reach it; they are using a different interpretation of “moving towards.”
Students choosing answer (2) may be confusing “origin” with “original position.”
For follow-up discussion, you can ask additional questions such as “Which people have a positive velocity
at t = 4 s?”
Question A2.06f
Description: Honing the concept of speed, and linking to graphical representations.
Question
During which 1-second time period(s) are there at least
two people with the same speed?
1. 0−1 s
2. 1−2 s
3. 2−3 s
4. 3−4 s
5. 4−5 s
6. 5−6 s
7. None of the time intervals
8. Two of the time intervals
9. Three of the time intervals
10. Four or more of the time intervals
Commentary
position (m)
1
2
Purpose: To hone your understanding of (instantaneous) speed.
Amy
Brad
t
Cate
2
1
0
x
0 1 2 3 4 5
time (s)
6
Discussion: The speed is the magnitude of velocity, and does not have a direction associated with it. An
object’s speed is the rate at which its position is changing at a particular instant of time—“how fast” it is
moving at that instant.
On a graph of position vs. time, velocity is the slope, and a straight line means that the velocity is constant.
So, for instance, during the fi rst second, Amy runs from x = −1 m to x = +2 m, a displacement of +3 m in
1 second, for a velocity of +3 m/s. Then, during the next 2 seconds, Amy walks from x = +2 m to x = +1 m,
a displacement of −1 m, for a velocity of 1 2 m s during that 2-second time interval. Meanwhile, Brad
has a velocity of 1 2 m s during the fi rst 2 seconds, so he and Amy have the same speed from t = 1 s
until t = 2 s.
During each 1-second time interval, each person’s speed is constant, so we can compare these speeds to
answer the question.
During three of the time intervals, two people have the same speed: (1) from t = 1 s to t = 2 s, both Amy and
Brad are moving at the same speed of 1 2 m s , though in opposite directions; (2) from t = 4 s to t = 5 s,
Amy and Cate are moving at the same speed, i.e., they are both not moving, so their speed is zero; and
(3) from t = 5 s to t = 6 s, Amy and Brad are again moving at the same speed, though this time it is 1 m/s.

Key Points:
• Speed is the magnitude of velocity.
• Two objects can have the same speed but different velocities, if they are moving in different directions.
Students who answer (8), two time intervals, and claim that the two intervals are 0–1 s and 2–3 s are may
be confusing position with speed, or interpreting the graph as velocity vs time. During these intervals, two
people cross paths. Students who interpret the plot as velocity vs. time may also have trouble identifying
intervals, since if this were such a graph Amy and Cate have the same “speed” at the instants t = 3 s and
t = 6 s.
Students choosing answer (8) for intervals 4−5 s and 5−6 s may be considering the direction of the velocity
as well as its magnitude, or may simply have overlooked that interval. Discussion should reveal this.
Students omitting interval 4−5 s may be thinking, explicitly or implicitly, that zero is not a speed.
Asking students to construct a sketch of speed vs. time may help them sort out the ideas here.
QUICK QUIZZES
1. (a) 200 yd (b) 0 (c) 0 (d) 8.00 yd/s
2. (a) False. The car may be slowing down, so that the direction of its acceleration is opposite the
direction of its velocity.
(b) True. If the velocity is in the direction chosen as negative, a positive acceleration causes a
decrease in speed.
(c) True. For an accelerating particle to stop at all, the velocity and acceleration must have opposite
signs, so that the speed is decreasing. If this is the case, the particle will eventually come to rest.
If the acceleration remains constant, however, the particle must begin to move again, opposite to
the direction of its original velocity. If the particle comes to rest and then stays at rest, the accel-eration
has become zero at the moment the motion stops. This is the case for a braking car—the
acceleration is negative and goes to zero as the car comes to rest.
3. The velocity vs. time graph (a) has a constant positive slope, indicating a constant positive
acceleration, which is represented by the acceleration vs. time graph (e).
Graph (b) represents an object whose speed always increases, and does so at an ever increasing
rate. Thus, the acceleration must be increasing, and the acceleration vs. time graph that best
indicates this behavior is (d).
Graph (c) depicts an object which fi rst has a velocity that increases at a constant rate, which
means that the object’s acceleration is constant. The motion then changes to one at constant
speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this
situation is graph (f ).

40 Chapter 2
4. (b). According to graph b, there are some instants in time when the object is simultaneously at
two different x-coordinates. This is physically impossible.
5. (a) The blue graph of Figure 2.14b best shows the puck’s position as a function of time. As seen in
Figure 2.14a, the distance the puck has traveled grows at an increasing rate for approximately
three time intervals, grows at a steady rate for about four time intervals, and then grows at a
diminishing rate for the last two intervals.
(b) The red graph of Figure 2.14c best illustrates the speed (distance traveled per time interval) of
the puck as a function of time. It shows the puck gaining speed for approximately three time
intervals, moving at constant speed for about four time intervals, then slowing to rest during the
last two intervals.
(c) The green graph of Figure 2.14d best shows the puck’s acceleration as a function of time.
The puck gains velocity (positive acceleration) for approximately three time intervals, moves
at constant velocity (zero acceleration) for about four time intervals, and then loses velocity
(negative acceleration) for roughly the last two time intervals.
6. (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its
acceleration is the free-fall acceleration, g = 9.80 m/s2.
7. (c). As it travels upward, its speed decreases by 9.80 m/s during each second of its motion. When
it reaches the peak of its motion, its speed becomes zero. As the ball moves downward, its speed
increases by 9.80 m/s each second.
8. (a) and (f ). The fi rst jumper will always be moving with a higher velocity than the second. Thus,
in a given time interval, the fi rst jumper covers more distance than the second, and the separa-tion
distance between them increases. At any given instant of time, the velocities of the jumpers
are defi nitely different, because one had a head start. In a time interval after this instant, how-ever,
each jumper increases his or her velocity by the same amount, because they have the same
acceleration. Thus, the difference in velocities stays the same.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1. Once the arrow has left the bow, it has a constant downward acceleration equal to the free-fall
acceleration, g. Taking upward as the positive direction, the elapsed time required for the velocity
to change from an initial value of 15.0 m/s upward (v0 = +15.0 m/s) to a value of 8.00 m/s
downwardward (vf = −8.00 m/s) is given by
v v v0 8 00 15 0
Δ Δ
t
f = =
−
−
a g
=
− . m s − (+ .
m s
)
−
9 .
80
m s2 = 2.35 s
Thus, the correct choice is (d).

2. The maximum height (where v = 0 ) reached by a freely falling object shot upward with an initial
= 2 + 2a(Δy) as
velocity v0 = +225 m s is found from v2 v
0
Δy
2 2 v m s
g
( ) =
− ( )
(− ) =
− ( )
0
2
0 225
2 980
0
max (− .
m s2 )= 2.58 × 103 m
Thus, the projectile will be at the Δy = 6.20 × 102 m level twice, once on the way upward and
once coming back down. The elapsed time when it passes this level coming downward can be
found by using Δy = v t − gt 0
1
2
2 and obtaining the largest of the two solutions to the resulting
quadratic equation:
. × 2 m = ( m s)t − (9.80 m s2 )t 2
6 20 10 225
1
2
or
(4.90 m s2 )t 2 − (225 m s)t + 6.20 × 102 m = 0
The quadratic formula yields
t =
−(−225 ) ± (−225 ) − 4 (4 90 ) 6 20 × 10 m s m s 2 . m s2 . 2 m
m s2
( )
2(4.90 )
with solutions of t = 43.0 s and t = 2.94 s . The projectile is at a height of 6.20 × 102 m and
coming downward at the largest of these two elapsed times, so the correct choice is seen to be (e).
3. The derivation of the equations of kinematics for an object moving in one dimension
( Equations 2.6, 2.9, and 2.10 in the textbook) was based on the assumption that the object had
a constant acceleration. Thus, (b) is the correct answer. An object having constant acceleration
would have constant velocity only if that acceleration had a value of zero, so (a) is not a neces-sary
condition. The speed (magnitude of the velocity) will increase in time only in cases when
the velocity is in the same direction as the constant acceleration, so (c) is not a correct response.
An object projected straight upward into the air has a constant acceleration. Yet its position
(altitude) does not always increase in time (it eventually starts to fall back downward) nor is its
velocity always directed downward (the direction of the constant acceleration). Thus, neither
(d) nor (e) can be correct.
4. The bowling pin has a constant downward acceleration (a = −g = −9.80 m s2 )while in fl ight.
The velocity of the pin is directed upward on the upward part of its fl ight and is directed down-ward
as it falls back toward the juggler’s hand. Thus, only (d) is a true statement.
5. The initial velocity of the car is v0 = 0 and the velocity at time t is v. The constant acceleration is
therefore given by a = Δv Δt = (v − v ) t = (v − ) t = v t 0 0 and the average velocity of the car is
v = (v + v ) = (v + ) = v 0 2 0 2 2. The distance traveled in time t is Δ x = vt = vt 2. In the special
case where a = 0 ( = = 0) 0 and hencev v , we see that statements (a), (b), (c), and (d) are all cor-rect.
However, in the general case (a ≠ 0, and hence v ≠ 0) only statements (b) and (c) are true.
Statement (e) is not true in either case.

42 Chapter 2
6. We take downward as the positive direction with y = 0 and t = 0 at the top of the cliff. The freely
falling pebble then has v0 = 0 and a = g = 9.8 m s2. The displacement of the pebble at t = 1.0 s
is given: y1 = 4.9 m. The displacement of the pebble at t = 3.0 s is found from
y t 3 0
1 0
2
2
1
2
= v + at 2 = + (9.8m s2 )(3.0 s) = 44 m
The distance fallen in the 2.0 s interval from t = 1.0 s to t = 3.0 s is then
Δy = y − y = − = 3 1 44 m 4.9 m 39 m
and choice (c) is seen to be the correct answer.
7. In a position vs. time graph, the velocity of the object at any point in time is the slope of the line
tangent to the graph at that instant in time. The speed of the particle at this point in time is simply
the magnitude (or absolute value) of the velocity at this instant in time. The displacement occur-ring
during a time interval is equal to the difference in x-coordinates at the fi nal and initial times
of the interval (Δx x x ) t f ti = − .
The average velocity during a time interval is the slope of the straight line connecting the points on
the curve corresponding to the initial and fi nal times of the interval [ v = Δx Δt = (x − x ) (t − t )] f i f i .
Thus, we see how the quantities in choices (a), (e), (c), and (d) can all be obtained from the graph.
Only the acceleration , choice (b), cannot be obtained from the position vs. time graph.
8. The elevator starts from rest v0 ( = 0) and reaches a speed of v = 6 m s after undergoing a
displacement of Δy = 30 m. The acceleration may be found using the kinematics equation
v2 v
= 2 + 2a(Δy) as
0
a
y
=
−
2 ( ) 2
−
( ) = v2 v
6 0
0
2
2 ( ) = 30
0 6
Δ
m s
m
. m s2
Thus, the correct choice is (c).
9. The distance an object moving at a uniform speed of v = 8 5. m s will travel during a time inter-val
of Δt = 1 1 000 s = 1.0 × 10−3 s is given by
Δx = v(Δt ) = (8.5 m s)(1.0 × 10−3 s) = 8.5 × 10−3 m = 8.5 mm
so the only correct answer to this question is choice (d).

10. Once either ball has left the student’s hand, it is a freely falling body with a constant acceleration
a = −g (taking upward as positive). Therefore, choice (e) cannot be true. The initial velocities of
the red and blue balls are given by v v v v iR iB = + = − 0 0 and , respectively. The velocity of either
ball when it has a displacement from the launch point of Δy = −h (where h is the height of the
building) is found from v2 = v2 + 2 ( ) i a y Δ as follows:
v = − v 2 + 2 a( Δ y) = − (+ v ) 2
+ 2 (−g)(−h) = − v 2 + 2
gh
R iR R 0
0
and
v = − v 2 + 2 a( Δ y) = − (− v ) 2
+ 2 (−g)(−h) = − v 2 + 2
gh
B iB B 0
0
Note that the negative sign was chosen for the radical in both cases since each ball is moving in
the downward direction immediately before it reaches the ground. From this, we see that choice
(c) is true. Also, the speeds of the two balls just before hitting the ground are
0 2 2 and v v v v B = − + gh = + gh 0
v = − v 2
+ gh = v 2
+ gh v R 0
0
2
2
2 2 .
0
0 Therefore, v v R B = , so both choices (a) and (b) are false. However, we see that both fi nal speeds
exceed the initial speed or choice (d) is true. The correct answer to this question is then (c) and (d).
11. At ground level, the displacement of the rock from its launch point is Δy = −h, where h is the
= 2 + 2a(Δy),
height of the tower and upward has been chosen as the positive direction. From v2 v
0
the speed of the rock just before hitting the ground is found to be
2 2 2a Δy 2 g h 12 m s 2(9.8 m s2 )(40.0 m) = 30 m s
v = ± v 2
+ ( ) = v + (− )(− ) = ( ) + 0
0
Choice (b) is therefore the correct response to this question.
12. Once the ball has left the thrower’s hand, it is a freely falling body with a constant, nonzero,
acceleration of a = −g. Since the acceleration of the ball is not zero at any point on its trajectory,
choices (a) through (d) are all false and the correct response is (e).
ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS
2. Yes. Zero velocity means that the object is at rest. If the object also has zero acceleration, the
velocity is not changing and the object will continue to be at rest.
4. No. They can be used only when the acceleration is constant. Yes. Zero is a constant.
6. (a) In Figure (c,) the images are farther apart for each successive time interval. The object is
moving toward the right and speeding up. This means that the acceleration is positive in (c).
(b) In Figure (a), the fi rst four images show an increasing distance traveled each time interval
and therefore a positive acceleration. However, after the fourth image, the spacing is
decreasing showing that the object is now slowing down (or has negative acceleration).
(c) In Figure (b), the images are equally spaced showing that the object moved the same
distance in each time interval. Hence, the velocity is constant in (b).

44 Chapter 2
8. (a) At the maximum height, the ball is momentarily at rest. (That is, it has zero velocity.) The
acceleration remains constant, with magnitude equal to the free-fall acceleration g and
directed downward. Thus, even though the velocity is momentarily zero, it continues to
change, and the ball will begin to gain speed in the downward direction.
(b) The acceleration of the ball remains constant in magnitude and direction throughout the ball’s
free fl ight, from the instant it leaves the hand until the instant just before it strikes the ground.
The acceleration is directed downward and has a magnitude equal to the free-fall acceleration g.
10. (a) Successive images on the fi lm will be separated by a constant distance if the ball has con-stant
velocity.
(b) Starting at the right-most image, the images will be getting closer together as one moves
toward the left.
(c) Starting at the right-most image, the images will be getting farther apart as one moves
toward the left.
(d) As one moves from left to right, the balls will fi rst get farther apart in each successive
image, then closer together when the ball begins to slow down.
PROBLEM SOLUTIONS
2.1 We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform
speed. The elapsed time is then
Δ Δ
t
= x = = × − =
v
2
2 10 2 0 02 m
100 m s
s . s
2.2 At constant speed, c = 3 × 108 m s, the distance light travels in 0.1 s is
8 7 1 m s s m
Δx = c(Δt ) = (3 × 10 )(0 1 ) = (3 × 10 )
mi
1.6
.
09 km
1
km
= × mi 3
10 m
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
2 104
Comparing this to the diameter of the Earth, DE, we fi nd
Δx Δ
D
= = ×
x
R E E
( × ) ≈
2
3 0 10
2 6 38 10
2 4
7
6
.
.
.
m
m
(with RE = Earth’s radius)
2.3 Distances traveled between pairs of cities are
Δx Δt 1 1 1 = v ( ) = (80.0 km h)(0.500 h) = 40.0 km
Δx Δt 2 2 2 = v ( ) = (100 km h)(0.200 h) = 20.0 km
Δ x Δt 3 3 3 = v ( ) = (40.0 km h)(0.750 h) = 30.0 km
Thus, the total distance traveled is Δx = (40.0 + 20.0 + 30.0) km = 90.0 km, and the elapsed
time is Δt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.
(a) v = = = Δ
Δ
x
t
90.0 km
1.70 h
52.9 km h
(b) Δx = 90.0 km (see above)

2.4 (a) v= = ⎛
⎝ ⎜
⎞
⎠ ⎟
×
Δ
Δ
x
t
20 ft 1 m
1
1 yr
3.281 ft
yr
m s 7
3.156 10 s
⎛
⎝ ⎜
⎞
⎠ ⎟
= 2 × 10−7
or in particularly windy times,
v= = ⎛
⎝ ⎜
⎞
⎠ ⎟
Δ
Δ
x
t
100 ft 1 m
1
1 yr
3.281 ft
yr
3.156 ×
⎛
⎝ ⎜
⎞
m s 7 1 10 6
⎠ ⎟
= × −
10 s
(b) The time required must have been
Δ Δ
t
x = = × ⎛
⎝ ⎜
⎞
⎠ ⎟
v
3 103 mi 1609 m
103
10 mm yr
1 mi
mm
1 m
yr
⎛
⎝ ⎜
⎞
⎠ ⎟
= 5 × 108
2.5 (a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h. Boat B requires
2.0 h to cross the lake at which time the race is over.
Boat A wins, being 60 km ahead of B when the race ends.
(b) Average velocity is the net displacement of the boat divided by the total elapsed time. The
winning boat is back where it started, its displacement thus being zero, yielding an average
velocity of zero .
2.6 The average velocity over any time interval is
v= =
−
−
Δ
Δ
x
t
x x
t t
f i
f i
(a) v= = −
Δ x
1 0.0 m
0
=
Δ
t
2.00 s
−
0
5.00 m s
(b) v= = −
Δ x
5 .00 m
0
=
Δ
t
4.00 s
−
0
1.25 m s
(c) v= = −
Δ x
5.00 m 10.0 m
= −
Δ
t
4.00 s −
2.00 s
2.50 m s
(d) v= =− −
Δ x
5.00 m 5.00 m
= −
Δ
t
7.00 s −
4.00 s
3.33 m s
(e) v= = −
Δ x
x x
−
2 1
= =
Δ
t
t −
t
−
2 1
0 0
8.00 s 0
0
1 h = Δx = (85 0 )(35 0 )
2.7 (a) Displacement km h min
km km ⎛⎝
60 0
. .
. min
⎞⎠
+ 130 = 180
(b) The total elapsed time is
Δt = + ( )⎛⎝ ⎜
⎞⎠ ⎟
1
h
35 0 15 0 +
. min . min 2.0
60.0 min
0 h = 2.84 h
so,
v = Δ = =
Δ
x
t
180
2 84
6
km
h
3.4 km h
.

46 Chapter 2
2.8 The average velocity over any time interval is
v= =
−
−
Δ
Δ
x
t
x x
t t
f i
f i
(a) v= = −
Δ x
4 0
= +
Δ
t
1 0 −
0
4 0
.
.
.
m 0
s
m s
(b) v= =− −
Δ x
2 0
= −
Δ
t
4 0 −
0
0 50
.
.
.
m 0
s
m s
(c) v= = −
Δ x
0 40
= −
Δ
t
5 0 −
1 0
1 0
.
. .
.
m
s s
m s
(d) v= = −
Δ x
0
0
=
Δ
t
5 0 s
−
0
0
.
2.9 The instantaneous velocity at any time is the slope of the
x vs. t graph at that time. We compute this slope by using
two points on a straight segment of the curve, one point on
each side of the point of interest.
(a) v 0 50
1 0 0
1 0 0
4
.
.
s .
s
s
.0 m
1.0 s
=
−
−
= = = x xt 4 0 . m s
(b) v 2
2 .5 s 10
s
.5 s s
.0 s 2 1 0
6.0 m
1.5 s
=
−
−
= − x x.
.
= − 4 0 . m s
(c) v 3
4 2
2
0
.0 s
.0 s .5 s
4.0 s .5 s 1.5 s
=
−
−
= =
x x
0
(d) v 4
5 .0 s 40
s
.0 s s
.5 s 5 4 0
2.0 m
1.0 s
=
−
−
= + x x.
.
= 2 0 . m s
2.10 (a) The time for a car to make the trip is t
x = Δ
v
. Thus, the difference in the times for the
two cars to complete the same 10 mile trip is
Δ Δ Δ
⎝ mi
⎞
⎜mi h
0 mi h x x = − = − = −
t t t
⎛
1 2
1 2
10
55
10
v v 7
mi
⎠ ⎟
⎛⎝
= 60 min
1 h 2.3 min
⎞⎠
(b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the
slower car in a time of 15.0 min. This distance is given by Δx Δt 1 1 = v ( ) = (55 mi h)(15 min).
The faster car pulls ahead of the slower car at a rate of
vrelative = 70 mi h − 55 mi h = 15 mi h
Thus, the time required for it to get distance Δx1 ahead is
Δ Δ
t
= ( )( )
1 55 15
x
=
mi h min
v 15 .
0
mi h relative
= 55 min
Finally, the distance the faster car has traveled during this time is
Δx Δt 2 2
1
= 70 mi h 55 min
h
60 min
v ( ) = ( )( )⎛⎝
⎞⎠
= 64 mi

2.11 The distance traveled by the space shuttle in one orbit is
( )
Circumference of Orbit = 2π r = 2π Earth’s radius + 200 miles
= π ( + ) = 2.61 × 104
2 3963 200 mi mi
Thus, the required time is
t = = × Circumference
average speed
. 4 mi
2 61 10
19 8
00
1 32
mi h
= . h
2.12 (a) v1
( )
= + 1
( ) = = +
1 1 1
Δ
Δ
x
t
L
t
L t
(b) v2
( )
= − 2
( ) = = −
2 2 2
Δ
Δ
x
t
L
t
L t
(c) vtotal
( )
( ) + ( )
= total
( ) = total
+
=
Δ
Δ
x Δ Δ
t
x x
t t
1 2
1 2
+ −
+
0
L L
=
= t t t +
t 1 2 1 2
0
. trip ( ) = (Δ ) =
(d) ave speed
total distance traveled
t
( Δx ) + ( Δx
)
+
= + +−
+
=
2
t t
+ total
L L
t t
L
t t
1 2
1 2 1 2 1 2
2.13 The total time for the trip is t = t + = t + 1 1 22 0 . min 0.367 h, where t1 is the time spent
traveling at v1 = 89.5 km h. Thus, the distance traveled is Δx = v t = v t 1 1 , which gives
Δx = (89 5 )t = (77 8 )(t + 0 367 ) = 77 8 1 1 . km h . km h . h ( . km h)t + km 1 28 5 .
or
89 5 77 8 28 5 1 ( . km h − . km h)t = . km
From which, t1 = 2.44 h for a total time of t = t + = 1 0.367 h 2.80 h .
Therefore, Δx = vt = (77.8 km h)(2.80 h) = 218 km .
2.14 (a) At the end of the race, the tortoise has been moving for time t and the hare for a time
t − 2.0 min = t − 120 s. The speed of the tortoise is vt = 0.100 m s, and the speed of the
hare is v v h t = 20 = 2.0 m s. The tortoise travels distance xt , which is 0.20 m larger than
the distance xh traveled by the hare. Hence,
x x t h = + 0.20 m
which becomes
v v t h t = (t − 120 s) + 0.20 m
or
(0.100 m s)t = (2.0 m s)(t − 120 s) + 0.20 m
This gives the time of the race as t = 1.3 × 102 s
(b) x t t t = v = (0.100 m s)(1.3 × 102 s) = 13 m

48 Chapter 2
2.15 The maximum allowed time to complete the trip is
ttotal = total distance
required average speed
=
⎛
= 1 600 m
250 km h
⎝ ⎜
⎞
⎠ ⎟
1 km h
0.278 m s
23.0 s
The time spent in the fi rst half of the trip is
t1
= = half distance
v
1
800 m
230 km h
1 km h
0.278 m s
s
⎛
⎝ ⎜
⎞
⎠ ⎟
= 12.5
Thus, the maximum time that can be spent on the second half of the trip is
t t t 2= − = total 1 23.0 s − 12.5 s = 10.5 s
and the required average speed on the second half is
v2
= half distance
= 800 m
= 1 km s
76.2 m s
t2 10 5 .
h
m s
= 274 km h
0.278
⎛
⎝ ⎜
⎞
⎠ ⎟
2.16 (a) In order for the trailing athlete to be able to catch the leader, his speed v1 ( ) must be greater
than that of the leading athlete v2 ( ), and the distance between the leading athlete and the
fi nish line must be great enough to give the trailing athlete suffi cient time to make up the
defi cient distance, d.
(b) During a time t the leading athlete will travel a distance d t 2 2 = v and the trailing athlete will
travel a distance d t 1 1 = v . Only when d d d 1 2 = + (where d is the initial distance the trail-ing
athlete was behind the leader) will the trailing athlete have caught the leader. Requir-ing
that this condition be satisfi ed gives the elapsed time required for the second athlete to
overtake the fi rst:
d d d t t d 1 2 1 2 = + or v = v +
giving
d − = =
v v
( − ) or
v v 1 2
1 2
t t d t
(c) In order for the trailing athlete to be able to at least tie for fi rst place, the initial distance
D between the leader and the fi nish line must be greater than or equal to the distance the
leader can travel in the time t calculated above (i.e., the time required to overtake the
leader). That is, we must require that
⎣ ⎢⎤
d ≥ = =
D d t
( − )
⎡
⎦ ⎥
2 2 2
1 2
v v
v v
or D
v
v v
≥
d 2
−
1 2
2.17 The instantaneous velocity at any time is the slope
of the x vs. t graph at that time. We compute this slope
by using two points on a straight segment of the curve,
one point on each side of the point of interest.
(a) vt= = −
10 0 m
0
2 00 s
0
= 1 00
5 00 .
−
.
.
. s
m s
(b) vt= = ( − )
5 . 00 10 .
0
4 . 00 2 .
00
m
s
( − ) = − 3 00
2 . 50 .
s
m s
(c) vt= = ( − )
. .
. . s
5 00 5 00
5 00 4 00
m
s
( − ) = 4 50
0 .
− (− )
( − ) = 7 50
(d) vt= =
0 500
.
8 00 7 00
5 . 00 .
s
. .
m
s
m s

2.18 (a) A few typical values are
t(s) x(m)
1.00 5.75
2.00 16.0
3.00 35.3
4.00 68.0
5.00 119
6.00 192
(b) We will use a 0.400 s interval centered at t = 4.00 s. We fi nd at t = 3.80 s, x = 60.2 m and
at t = 4.20 s, x = 76.6 m. Therefore,
v = Δ = =
Δ
x
t
16 4. m
0.400 s
41 0 . m s
Using a time interval of 0.200 s, we fi nd the corresponding values to be: at t = 3.90 s,
x = 64.0 m and at t = 4.10 s, x = 72.2 m. Thus,
v = Δ = =
Δ
x
t
8.20 m
0.200 s
41 0 . m s
For a time interval of 0.100 s, the values are: at t = 3.95 s, x = 66.0 m, and at t = 4.05 s,
x = 70 1 . m. Therefore,
v = = = Δ
Δ
x
t
4.10 m
0.100 s
41 0 . m s
(c) At t = 4.00 s, x = 68.0 m. Thus, for the fi rst 4.00 s,
v= = −
Δ x
6 8.0 m
0
=
Δ
t
4.00 s
−
0
17 0 . m s
This value is much less than the instantaneous velocity at t = 4.00 s.
2.19 Choose a coordinate axis with the origin at the fl agpole and east as the positive direction. Then,
using x = x + t + at 0 0
v 2 with a = 0 for each runner, the x-coordinate of each runner at time t is
1
2
x t x A B = − 4.0 mi + (6.0 mi h) and = 3.0 mi + (− 5.0 mi h)t
When the runners meet, x x A B = , giving − 4.0 mi + (6.0 mi h)t = 3.0 mi + (− 5.0 mi h)t,
or (6.0 mi h + 5.0 mi h)t = 3.0 mi + 4.0 mi. This gives the elapsed time when they meet as
t= = 7 0
0 64
.
.
mi
11.0 mi h
h
At this time, x x A B = = −0.18 mi. Thus, they meet 0.18 mi west of the flagpole .

50 Chapter 2
2.20 (a) Using v = v + 0 at with an initial velocity of v0 = 13.0 m s and a constant acceleration of
a = −4.00 m s2, the velocity after an elapsed time of t = 1.00 s is
v = v + = + (− )( ) = 0 at 13.0 m s 4.00 m s2 1.00 s 9.00 m s
(b) At an elapsed time of t = 2.00 s, v = 13.0 m s + (−4.00 m s2 )(2.00 s) = 5.00 m s .
(c) When t = 2.50 s, v = 13.0 m s + (−4.00 m s2 )(2.50 s) = 3.00 m s .
(d) At t = 4.00 s, v = 13.0 m s + (−4.00 m s2 )(4.00 s) = −3.00 m s .
(e) The graph of velocity versus time for this canister is a straight line passing through
13.0 m s at t = 0 and sloping downward, decreasing by 4.00 m s for each second
thereafter.
(f ) If the canister’s velocity at time t = 0 and the value of its (constant) acceleration are known,
one can predict the velocity of the canister at any later time.
2.21 The average speed during a time interval is
v = distance traveled
Δt
During any quarter mile segment, the distance traveled is
= 1 5 280
Δx = ⎛⎝
⎞⎠
1 320
mi
4
ft
1 mi
ft
(a) During the fi rst quarter mile segment, Secretariat’s average speed was
v1
1 320
= = 52 4 ft
25.2 s
. ft s
During the second quarter mile segment,
v2
1 320
= = 55 0 ft
24.0 s
. ft s
For the third quarter mile of the race,
v3
1 320
= = 55 5 ft
23.8 s
. ft s
For the fourth fi nal quarter mile,
v4
1 320
= = 56 9 ft
23.2 s
. ft s
and during the fi nal quarter mile,
v5
1 320
= = 57 4 ft
23.0 s
. ft s
(b) Assuming that v v final = 5 and recognizing that v0 = 0, the average acceleration for the
entire race was
a =
−
v v = − final ft s
0 57 4 0
25.
total elapsed time
.
2 24.0 23.8 23.2 23.0 s
ft s2
( + + + + ) = 0.481

2.22 From a = Δv Δt, the required time is seen to be
g = = − ⎛
Δ Δ
t
⎝ ⎜
a g
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
v 60 0 0
7
1
9 8
.
.
mi h
0 m s
0.
2
447 m s
mi h
s
1
0 391
⎛
⎝ ⎜
⎞
⎠ ⎟
= .
= = ( − ) ⎛
2.23 From a = Δv Δt, we have Δ Δ
t
v 60 55 ⎞
a
⎝ ⎜
0 60
.
mi h 0 447
m s
m s
. 2 1 mi h
⎠ ⎟
= 3 7 . s .
2.24 (a) From t = 0 to t = 5 0 . s,
a
f i
t t
f i
=
−
−
=
− m s − (− m s
)
−
=
v v 8 0 8 0
5 0 0
0
. .
.
s
From to t = 15 s,
a =
− (− )
−
=
8 0 8 0
1 50
1 6
. .
.
.
m s m s
5 s s
m s2
and from t = 0 to t = 20 s,
a =
− (− )
−
=
8 0 8 0
2 0
0 80
. .
.
m s m s
0 s
m s2
(b) At any instant, the instantaneous acceleration equals the slope of the line tangent to the v vs. t
graph at that point in time. At t = 2 0 . s, the slope of the tangent line to the curve is 0 .
At t = 10 s, the slope of the tangent line is 1 6 . m s2 , and at t = 18 s, the slope of the
tangent line is 0 .
2.25 (a) a
= Δ v = 175 mi h
− 0
= 70 0
⋅
Δ
t
2 5
s
mi h s
.
.
or
a =
⋅
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
1609
1
⎠ ⎟70 0
1
.
mi
h s
m
mi
h
3 600 s
⎛
⎞
⎟
⎠ ⎜⎝ = 31 3. m s2
Alternatively,
a
31 3 =
3 19 2 .
s
g
g =⎛⎝
⎞⎠
⎛
⎝ ⎜
⎞
⎠ ⎟
1
.
9 80
.
m
m s2
(b) If the acceleration is constant, Δx = v t + 1
at 0
2
2:
1
2
Δx= + ⎛⎝
⎞⎠
m
s
0 ( ) =
31 3 2 50 97 8 2 . . .
s m 2
or
Δx = ( )⎛⎝
⎞⎠
3 .
281
97 8 =
. m
321
ft
1 m
ft

52 Chapter 2
2.26 We choose eastward as the positive direction so the initial velocity of the car is given by
v0 = +25.0 m s.
(a) In this case, the acceleration is a = +0.750 m s2 and the fi nal velocity will be
v = v + = + + (+ )( ) = + 0 at 25.0 m s 0.750 m s2 8.50 s 31.4 m s
or
v = 31.4 m s eastward
(b) When the acceleration is directed westward, a = −0.750 m s2, the fi nal velocity is
v = v + = + + (− )( ) = + 0 at 25.0 m s 0.750 m s2 8.50 s 18.6 m s, or v = 18.6 m s eastward .
2.27 Choose the direction of the car’s motion (eastward) as the positive direction. Then, the initial
velocity of the car is v0 = +40.0 m s and the fi nal velocity (after an elapsed time of Δt = 3.50 s)
is v = +25 0. ms.
(a) The car’s acceleration is
a
Δ v v v= 25 0 − 40 0
0 = −
Δ Δ
= =
−
t t
3 50
4 29
. .
.
.
m s m s
s
m s2 or a = 4.29 m s2 westward
(b) The distance traveled during the 3.50 s time interval is
+ ⎛⎝
⎞⎠
Δx = (Δt ) = Δt
= + ⎛⎝
⎞⎠
v
v v
av
25 . 0 m s 40 .
0
m s 0
2
2
(3.50 s) = 114 m
= 2 + 2 a(Δx), we have 10 97 103 2 0 2 220 ( . × m s) = + a( m) so that
2.28 From v2 v
0
a
x
=
2 3 2
−
( ) =
( × ) −
v2 v
0
= 2
( ) 10 97 10 0
2 220
2 7
Δ
.
.
m s
m
4 10
2 74 10
1
5
5
×
= ( × )⎛
⎝ ⎜
⎞
⎠ ⎟=
m s
m s
g
9.80 m s
2
2
2 .
2.79 × 104 times g !
2.29 (a) Δx = v (Δt ) = (v + v )Δt av 0 2 becomes 40 0
2 80
m s
+ ⎛⎝
v ( )
⎞⎠
= s
. 0 8 50
2
.
m .
,
which yields v0
2
8 50
= (40 0 ) − 2 80 = 6 61
.
. . .
s
m ms ms
(b) a
t
=
−
= − = − v v0 2 80 6 61
8 50
0 448
Δ
. .
.
.
m s m s
s
m s2
2.30 (a)
(b) The known quantities are initial velocity, fi nal velocity, and displacement. The kinematics
equation that relates these quantities to acceleration is v v f i 2 = 2 + 2a(Δx) .
continued on next page

(c) a
−
( )
v2 v2
2 Δ
f i =
x
(d) a
−
( ) = ( ) − ( )
v2 v2 2 2
2
f i =
. .
m s m s
m s2 ( ) = .
Δ x
2 2 00 1
30 0 20 0
×
.
0
1 25 2 m
= − = v v v 30 0 20 0
(e) Using a = Δv Δt, we fi nd that Δ Δ
t
−
f i = =
a a
1 25
8 00
. .
.
.
m s m s
m s2 s .
= 2 + 2a(Δx) yields
2.31 (a) With v = 120 km h, v2 v
0
a
x
=
−
( ) =
( ) − ⎡⎣
⎤⎦
( )
v2 v
2
0
2
2
120 0
2 240
⎛
0 278
2 Δ
km h
m
. m s
1 km h
⎝ ⎜
⎞
⎠ ⎟
= 2.32 m s2
(b) The required time is Δt
a
=
−
=
( − ) ⎛
km h 0 .
278
m s
⎜
⎝
v v0 120 0
2 32
m s
. 2 1 km h
⎞
⎠ ⎟
= 14 4. s .
2.32 (a) The time for the truck to reach 20 m s, starting from rest, is found from v = v + 0 at:
t
−
= − = v v0 20 0
speed
up
a 2
m s
m s
= s
2 0
10
.
braking = + + = 10 s + 20 s + 5.0 s = 35 s .
The total time for the trip is t t t t total speed
up
constant
speed
(b) The distance traveled during the fi rst 10 s is
(Δx) = t = t + ⎛⎝ ⎜
⎞⎠ ⎟
speed
up
sp v
speed
up
speed
up
v v0
2 eed
up
= 20 m s + 0
⎜
s m ⎛⎝ ⎞⎠ ⎟
( ) =
2
10 100
The distance traveled during the next 20 s (with a = 0) is
(Δx)constant = ⋅ t = ( )
speed
v 20 m s (20 s) = 400 m
constant
speed
The distance traveled in the last 5.0 s is
⎛ +
v v
2 aking
Δx t t ( ) = = f
braking braking braking br v
⎝ ⎜
⎞
⎠ ⎟
= 0 + ⎛⎝
20ms
.0 s m ⎞⎠
( ) =
2
5 50
The total displacement is then
(Δx) = (Δx) + (Δx) + (Δx total speed
up
constant
speed
) = + + = braking 100 m 400 m 50 m 550 m
and the average velocity for the entire trip is
vtrip
( ) = total
= =
m s total
m
35 s
Δx
t
550
16

54 Chapter 2
2.33 (a) a
t
=
−
= − = v v0 24 0 0
2 95
8 14
Δ
.
.
.
m s
s
m s
2
2
(b) From a = Δv Δt , the required time is Δt
−
v v 20 0 10 0
f i =
a
= − =
8 14
1 23
. .
.
.
m s m s
m s
s 2 .
(c) Yes. For uniform acceleration, the change in velocity Δv generated in time Δt is given by
Δv = a(Δt ). From this, it is seen that doubling the length of the time interval Δt will always
double the change in velocity Δv. A more precise way of stating this is: “When acceleration
is constant, velocity is a linear function of time.”
2.34 (a) The time required to stop the plane is
t
a
=
−
= −
−
= v v0 0 100 m s
5.00 m s2 20 0 . s
(b) The minimum distance needed to stop is
+ ⎛⎝
⎞⎠
⋅ = + ⎛⎝
v v0
2
Δx = ⋅ t = t
⎞⎠
0 100
v ( ) =
20 0 10
m s
2
. s 00 m =1.00 km
Thus, the plane requires a minimum runway length of 1.00 km.
It cannot land safely on a 0.800 km runway .
2.35 We choose x = 0 and t = 0 at location of Sue’s car when she fi rst spots the van and applies the
brakes. Then, the initial conditions for Sue’s car are x 0S = 0 and v0 30 0 S = . ms. Her
constant acceleration for t aS ≥ 0 is = −2.00 m s2. The initial conditions for the van are
x0 0 155 5 00 V V = m, v = . m s and its constant acceleration is aV = 0. We then use
Δx = x − x = t + at 0 0
v 2 to write an equation for the x coordinate of each vehicle for t ≥ 0.
1
2
This gives
Sue’s Car:
x t t S− 0 = (30 0 ) + (− )
1
2
. m s 2.00 m s2 2 or x t t S = (30.0 m s) − (1.00 m s2 ) 2
Van:
1
2
x t t V− 155 = (5 00 ) + ( )
m . m s 0 2 or x m m s t V= 155 + (5.00 )
In order for a collision to occur, the two vehicles must be at the same location i.e., x x S V ( = ).
Thus, we test for a collision by equating the two equations for the x-coordinates and see if the
resulting equation has any real solutions.
x x S V = ⇒ (30.0 m s)t − (1.00 m s2 )t 2 = 155 m + (5.00 m s)t
or (1.00 m s2 )t 2 − (25.00 m s) + 155 m = 0
Using the quadratic formula yields
t =
−(−25 00 ) ± (−25 00 ) − 4 (1 00 ) 155 2 . m s . m s . m s2 m
( ) =
2 ms
( )
s or 11.4 s 2
1 00
13 6
.
.
The solutions are real, not imaginary, so a collision will occur . The smaller of the two solutions
is the collision time. (The larger solution tells when the van would pull ahead of the car again
if the vehicles could pass harmlessly through each other.) The x-coordinate where the collision
occurs is given by
x x x collision S t s V t s = = = m + =11 4 =11 4 155 5 00 . . ( . m s)(11.4 s) = 212 m

2.36 The velocity at the end of the fi rst interval is
v = v + = + ( ) = 0 at 0 (2.77 m s) 15.0 s 41.6 m s
This is also the constant velocity during the second interval and the initial velocity for the third
interval.
1 2
2
(a) From Δx = v t + at 0
, the total displacement is
(Δx) = (Δx) + (Δx) + (Δx)
1 2 3
= + ( )
total
0
m s2
( ) ⎡⎣⎢
1
2
2.77 15 0 4 1 0
4
2 . s 1.6 m s 23 s
1
⎤⎦⎥
+ ( )( ) + ⎡⎣
⎤⎦
1
2
+ .6 m s .39 s 9.47 m s .39 s 2 ( )( ) + − ( )( ) ⎡⎣⎢
4 ⎤
4 2
⎦⎥
or
(Δx) = + × + = × total 312 m 5.11 103 m 91.2 m 5.51 103 m = 5.51 km
(b) v
1
( = )= 1
=
20 8 1
312
Δx
t
m
15.0 s
. m s
v
2
( ) = × 2
= =
41 6 2
5
Δx
t
.11 10 m
123 s
m s
3
.
v
3
( = )= 9 3
=
Δx
t
3
1.2 m
4.39 s
20 8 . m s
and the average velocity for the total trip is
vtotal
( ) = = ×
m total
total
. 3
. .
( ) =
38 7 + +
Δx
t
5 51 10
15 0 123 4
.
39
s
m s
2.37 Using the uniformly accelerated motion equation Δx = v t + 1
at 0
2
2 for the full 40 s interval yields
Δx = (20 )(40 ) + 1 (−1 0 )(40 ) = 0
2 m s s . m s2 s , which is obviously wrong.
2
The source of the error is found by computing the time required for the train to come to rest. This
time is
t
a
=
−
= −
−
= v v0 0 20
20
m s
1.0 m s
s 2
Thus, the train is slowing down for the fi rst 20 s and is at rest for the last 20 s of the 40 s interval.
The acceleration is not constant during the full 40 s. It is, however, constant during the fi rst
20 s as the train slows to rest. Application of Δx = v t + 1
at 0
2
2 to this interval gives the stopping
distance as
Δx = (20 )(20 ) + (−1 0 )(20 ) = 1
2 m s s . m s2 s 200 m
2

56 Chapter 2
0 447 = =⎛⎝
2.38 v v 0 0 40 0
⎞⎠
⎛
⎝
and
mi
h
m s
.
⎜
1 mi h f .
⎞
⎠ ⎟
= 17 9. m s
(a) To fi nd the distance traveled, we use
⎛ +
v v0
Δx t t f = ⋅ =
⎝ ⎜
⎞
⎠ ⎟
⋅ = + ⎛⎝
⎞⎠
v (
2
17 9 0
2
12 0
.
.
m s
s) = 107 m
(b) The constant acceleration is
a
v v0 17 9 0
t
f =
−
= − =
12 0
1 49
.
.
.
m s
s
m s2
2.39 At the end of the acceleration period, the velocity is
v = v + = + ( )( ) = 0 at 0 1 5 5 0 7 5 accel
. m s2 . s . m s
This is also the initial velocity for the braking period.
(a) After braking, v v f = + at = + (− )( ) = brake
7.5 m s 2.0 m s2 3.0 s 1.5 m s .
(b) The total distance traveled is
Δx Δx Δx t t total accel brake accel = ( ) + ( ) = (v ⋅ ) + (v ⋅ ) =
v v v v 0
2 2
+ ⎛⎝
⎞⎠
t t f
+
⎛ +
⎝ ⎜
⎞
⎠ ⎟
brake accel brake
Δxtotal
= 7 5 m s
+ 0 ( ) + m s +
m s s
⎛⎝
⎞⎠
2
5 0
1 5 7 5
2
.
.
. . ⎛⎝
⎞⎠
(3.0 s) = 32 m
2.40 For the acceleration period, the parameters for the car are: initial velocity = v = ia 0,
acceleration = a = a a 1, elapsed time = (Δt ) = t a 1, and final velocity = vfa. For the braking period,
the parameters are: initial velocity = v = final vel. of accel. p ib eriod = vfa, acceleration = a = a b 2,
and elapsed time = (Δt ) = t b 2 .
(a) To determine the velocity of the car just before the brakes are engaged, we apply
v v f i = + a(Δt ) to the acceleration period and fi nd
v v v ib fa ia a a = = + a (Δt ) = 0 + a t 1 1 or vib = a t 1 1
(b) We may use Δx = v ( Δt ) + 1 a ( Δt )
i 2
2 to determine the distance traveled during the
acceleration period (i.e., before the driver begins to brake). This gives
( Δx ) = v ( Δt ) + 1 a ( Δt ) 2
= +
a t a ia a 2
a a 0 2
1 1
1
2
or Δx at a ( ) = 1
2
2 1 1
(c) The displacement occurring during the braking period is
1 2
2
( Δx ) = v ( Δt ) + 1 a ( Δt ) 2
= ( a t ) t +
a t b ib b 2
b b 1 1 2 2 2
Thus, the total displacement of the car during the two intervals combined is
1
2
1
2 1 1
( Δx ) = ( Δx ) + ( Δx )= a t 2
+ a t t + a t total
a b 2
1 1 2 2 2

2.41 The time the Thunderbird spends slowing down is
= = 6 99 ( )
Δ Δ Δ
t
2 2 250
x x
1
1
1
1
0
0 715
+
= ( )
+
=
v v v
m
. m s
. s
The time required to regain speed after the pit stop is
= = 9 79 ( )
Δ Δ Δ
t
2 2 350
x x
2
2
2
2
0
71 5 0
+
= ( )
+
=
v v v
m
. m s
. s
Thus, the total elapsed time before the Thunderbird is back up to speed is
Δt = Δt + + Δt = + + = 1 2 5.00 s 6.99 s 5.00 s 9.79 s 21.8 s
During this time, the Mercedes has traveled (at constant speed) a distance
Δx Δt M = v ( ) = ( )( ) = 0 71.5 m s 21.8 s 1 558 m
and the Thunderbird has fallen behind a distance
d x x x M = Δ − Δ − Δ = − − 1 2 1 558 m 250 m 350 m = 958 m
2.42 The car is distance d from the dog and has initial velocity v0 when the brakes are applied, giving
it a constant acceleration a.
Apply v = Δx Δt = v + v0 2 to the entire trip (for which Δx = d + 4.0 m, Δt = 10 s, and v = 0)
to obtain
4 0 0 +
d + =
. m
0 10 s
2
v
or v0
4 0 = d + . m
5.0 s
[1]
= 2 + 2a(Δx) to the entire trip yields 0 2 4 0 0
Then, applying v2 v
0
= v2 + a(d + . m).
Substitute for v0 from Equation [1] to fi nd that
0
4 0
2 40
2
=
( d
+ .
m
)+ a ( d
+ .
) 25 s
m 2 and a
d = − + 4 0. m
50 s2 [2]
Finally, apply Δx = v t + 1
at 0
2
2 to the fi rst 8.0 s of the trip (for which Δx = d).
This gives
d = v ( ) + a( ) 0
1
2 8.0 s 64 s2 [3]
Substitute Equations [1] and [2] into Equation [3] to obtain
d
d d = + ⎛
⎝ ⎜
⎞
50 s2 ⎟ ( ) = + 64 s2 0.96d 3.84 m
. m 4 .
0
⎠ ⎟
( ) + ⎛ − +
⎝ ⎜
⎞
⎠
4 0
8 0
1
2
.
5.0 s
s
m
which yields d = 3.84 m 0.04 = 96 m .

58 Chapter 2
2.43 (a) Take t = 0 at the time when the player starts to chase his opponent. At this time, the
opponent is distance d = (12 m s)(3.0 s) = 36 m in front of the player. At time t 0, the
displacements of the players from their initial positions are
1 2
2
Δx = (v ) t + a t = 0
+ .0 m s player 0
player player 1
2
(4 2 )t 2 [1]
and
1 2
2
Δx t a t opponent opponent opponent = (v ) + = m 0
(12 s)t + 0 [2]
When the players are side-by-side,
Δx Δx player opponent = + 36 m [3]
Substituting Equations [1] and [2] into Equation [3] gives
1
2
(4.0 m s2 )t 2 = (12 m s)t + 36 m or t 2 + (− 6.0 s)t + (−18 s2 ) = 0
Applying the quadratic formula to this result gives
t =
−(− 6 . 0 s ) ± (− 6 . 0 s ) 2 − 4 ( 1 )(− 18
s2
)
( )
2 1
which has solutions of t =−2 2 . s and t =+8 2 . s. Since the time must be greater than zero,
we must choose t = 8 2 . s as the proper answer.
1 2
2
(b) Δx = (v ) t + a t = 0
+ .0 m s player 0
player player 1
2
(4 2 )(8 2 s) = 2 . 1.3 × 102 m
2.44 The initial velocity of the train is v0 = 82.4 km h and the fi nal velocity is v = 16.4 km h. The
time required for the 400 m train to pass the crossing is found from
Δx t t = ⋅ = + ( ) ⎡⎣
⎤⎦
2 as
v v v0
t
x = ( )
+
= ( )
( + ) =
2 2 0 400
82 4 16 4
8 10
0
Δ
v v
.
. .
.
km
km h
× ( )⎛⎝
⎞⎠
3 600
s
10− =
3 h 29 1
1 h
. s
= 2 + 2a(Δy) with v = 0, we have
2.45 (a) From v2 v
0
Δy
− ( )
= max a
(− 2
) ( ) =
−
2 2
=
m s
m s
v2 v
0
2
0 250
.
2 980
.
31 9 . m
(b) The time to reach the highest point is
t
m s
a up 2
9.80 m s
=
−
= −
−
= v v0 0 25.0
2.55 s
(c) The time required for the ball to fall 31.9 m, starting from rest, is found from
1
2
Δy = (0)t + at
2 as t
= ( )=
y
a
(− )
−
=
2 2 319
9 80
2 55
Δ .
.
.
m
m s
s 2
(d) The velocity of the ball when it returns to the original level (2.55 s after it starts to fall from
rest) is
v = v + = + (− )( ) = 0 at 0 9.80 m s2 2.55 s
− 25 0 . m s

2.46 (a) For the upward fl ight of the arrow, v0 = +100 m s, a = −g = −9.80 m s2, and the fi nal
= 2 + 2a(Δy) yields
velocity is v = 0. Thus, v2 v
0
Δy
− ( )
= max a
(− .
) ( ) =
−
2 2
=
v2 v
0
2
0 100
2 980
m s
m s2 510 m
(b) The time for the upward fl ight is
t
y y
up
up o
m
m s
= ( ) = ( )
+
= ( )
+
=
Δ Δ max max
v v v
2 2 510
100 0
10 2. s
For the downward fl ight, Δy = −(Δy) = − = a = − max 510 0, 9.8 0 m, v and m s2 . Thus,
Δy = v t + 1
at 0
2
2 gives t
= s ( ) =
(− m
)
−
m s
y
a down 2
=
2 2 510
9 80
10 2
Δ
.
.
and the total time of the fl ight is t t t total down down = + = 10.2 s + 10.2 s = 20.4 s .
2.47 The velocity of the object when it was 30 0. m above the ground can be determined by applying
Δy = v t + 1
at 0
2
2 to the last 1.50 s of the fall. This gives
1
2
− = ( ) + − ⎛⎝
⎞⎠
m
s
30 0 1 50 ( )
2 . m . s . .
v 9 80 1 50 s or v= −12.7 m s
0
2 0 The displacement the object must have undergone, starting from rest, to achieve this velocity at a
point 30.0 m above the ground is given by v2 v
= 2 + 2a(Δy) as
0
Δy
(− ) −
(− ) = − 1
a
( ) =
−
2 2
=
2
0
2
12 7 0
2 980
m s
8 2
3 m
m s2 v v .
.
.
The total distance the object drops during the fall is
(Δy) = (Δy) + (− ) = total m m 1 30.0 38.2
2.48 (a) Consider the rock’s entire upward fl ight, for which v0 = +7.40 m s,
vf = 0, a = −g = −9.80 m s2, yi = 1.55 m (taking y = 0 at ground level), and
y h f= = max maximum altitude reached by rock . Then applying v v f i 2 = 2 + 2a(Δy) to this
upward fl ight gives
0 7 40 2 9 80 1 55 2 = ( . ) + (− . )( − . ) max m s m s2 h m
and solving for the maximum altitude of the rock gives
= + . ( )
hmax .
7 .
40
2 9 .
80
( ) 1 55 =
4 34
2
m
m s
m s
m 2
Since hmax 3.65 m (height of the wall), the rock does reach the top of the wall .
(b) To fi nd the velocity of the rock when it reaches the top of the wall, we use v v f i 2 = 2 + 2a(Δy)
and solve for vf when yf = 3.65 m (starting with vi i= +7.40 m s at y = 1.55 m). This
yields
v v f i f i = 2 + a(y − y ) = ( )2 + (− ) 2 7.40 m s 2 9.80 m s2 (3.65 m − 1.55 m) = 3.69 m s

60 Chapter 2
(c) A rock thrown downward at a speed of 7.40 m s v 7.40 m s i ( = − )from the top of the wall
undergoes a displacement of Δy y y f i ( ) = − = 1.55 m − 3.65 m = −2.10 m
before reaching
the level of the attacker. Its velocity when it reaches the attacker is
v v f i = − 2 + a( y) = − (− )2 + (− ) − 2 Δ 7.40 m s 2 9.80 m s2 ( 2.10 m) = −9.79 m s
so the change in speed of this rock as it goes between the 2 points located at the top of the
wall and the attacker is given by
Δ speed f i ( ) = − =− −− = down v v 9.79 m s 7.40 m s 2.39 m s
(d) Observe that the change in speed of the ball thrown upward as it went from the attacker to
the top of the wall was
Δ speed f i ( ) = − = − = up v v 3.69 m s 7.40 m s 3.71 m s
Thus, the two rocks do not undergo the same magnitude change in speeds . As the two
rocks travel between the level of the attacker and the level of the top of the wall, the rock
thrown upward undergoes a greater change in speed than does the rock thrown downward.
The reason for this is that the rock thrown upward has a smaller average speed between
these two levels:
v
v v
up
up up 7.40 m s m s
= m s
2
+
= + = i f
2
3 69
5 55
.
.
and
v
v v
down
down down 7.40 m s m s
2
=
+
= + = i f
2
9 79
8 60
.
. m s
Thus, the rock thrown upward spends more time traveling between the two levels ,
with
gravity changing its speed by 9.80 m s for each second that passes.

2.49 The velocity of the child’s head just before impact (after falling a distance of 0.40 m, starting
= 2 + 2a(Δy) as
from rest) is given by v2 v
0
v v I = − + a( y) = − + (− )(− ) = − 0
2 2 Δ 0 2 9.8 m s2 0.40 m 2.8 m s
If, upon impact, the child’s head undergoes an additional displacement Δy = −h before
coming to rest, the acceleration during the impact can be found from v2 v
= 2 + 2a(Δy)
0
to
be a h h I I = (0 − v2 ) 2(− ) = v2 2 . The duration of the impact is found from v = v + 0 at as
t a h I I = Δv = −v (v2 2 ), or t h I = −2 v .
Applying these results to the two cases yields:
Hardwood Floor (h = 2.0 × 10−3 m):
a
= I =
h
(− )
( × ) = −
v2 2
2 3
.
.
2 8
m s
2 2 0 10
m
2.0 × 103 m s2
and t
= − h
=
I
− ( × )
−
= × =
−
2 2 2 0 10 −
2 8
1 4 10
3
3
v
.
.
.
m
m s
s 1 4 . ms
Carpeted Floor ( h = 1.0 × 10−2 m ):
a
= I =
h
(− )
( × ) = −
v2 2
2 2
.
.
2 8
m s
2 1 0 10
m
3.9 × 102 m s2
and t
= − h
=
I
− ( × )
−
= × =
−
2 2 1 0 10 −
2 8
7 1 10
2
3
v
.
.
.
m
m s
s 7 1 . ms
2.50 (a) After 2.00 s, the velocity of the mailbag is
= + = − m s + (− m s2 )( s) = − 0 at 1.50 9.80 2.00 21.1 m s
v v bag
The negative sign tells that the bag is moving downward and the magnitude of the velocity
gives the speed as 21 1 . m s .
(b) The displacement of the mailbag after 2.00 s is
+ ⎛⎝
⎞⎠
(Δy) = t
=
⎡− + (− )
⎣ ⎢⎤
bag
v v 21 . 1 m s 1 .
50
m s 0
2
2
⎦ ⎥
(2.00 s) = − 22.6 m
During this time, the helicopter, moving downward with constant velocity, undergoes a
displacement of
1 2
2
(Δy) = t + at = (− )( ) + = − copter v m s s 0
1.5 2.00 0 3.00 m
The distance separating the package and the helicopter at this time is then
d y y p h = (Δ ) − (Δ ) = −22.6 m − (−3.00 m) = −19.6 m = 19.6 m

62 Chapter 2
(c) Here, (v0 ) = (v0 ) = +1 50 bag copter . m s and abag
= −9.80 m s2 while acopter = 0. After 2.00 s,
the velocity of the mailbag is
= m
+ − m
s
⎛⎝
s m
s
vbag 2
s
⎞⎠
1.50 9.80 (2.00 ) = −18.1
and its speed is
vbag
m
s = 18 1 .
In this case, the displacement of the helicopter during the 2.00 s interval is
Δycopter = (+1.50 m s)(2.00 s) + 0 = +3.00 m
Meanwhile, the mailbag has a displacement of
⎛ +
(Δy) = t
⎝ ⎜
⎞
⎠ ⎟
= − + ⎡⎣
bag
vbag v m s m s 0
2
18 . 1 1 .
50
2
⎢
⎤⎦⎥
(2.00 s) = −16.6 m
The distance separating the package and the helicopter at this time is then
d y y p h = (Δ ) − (Δ ) = −16.6 m − (+3.00 m) = −19.6 m = 19.6 m
2.51 (a) From the instant the ball leaves the player’s hand until it is caught, the ball is a freely
falling body with an acceleration of
a = −g = −9.80 m s2 = 9.80 m s2 (downward)
(b) At its maximum height, the ball comes to rest momentarily and then begins to fall back
downward. Thus, vmax
height
= 0 .
(c) Consider the relation Δy = v t + 1
at 0
2
2 with a = − g. When the ball is at the thrower’s hand,
= v t − gt 2
the displacement is Δy = 0, giving 0 0
1
2
This equation has two solutions, t = 0 which corresponds to when the ball was thrown,
and t = 2 g 0 v corresponding to when the ball is caught. Therefore, if the ball is caught at
t = 2.00 s, the initial velocity must have been
= =( )( )
= gt . m s2 . s
v0 2
9 80 2 00
2
9.80 m s
= 2 + 2a(Δy), with v = 0 at the maximum height,
(d) From v2 v
0
Δy
− ( )
= max a
(− 2
) ( ) =
−
2 2
=
m s
m s
v2 v
0
2
0 980
.
2 980
.
4.90 m

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