Solucionario Fundamentos de Física 9na edición Capitulo 2

2Motion 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.

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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. 12 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, − 12 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.


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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 10. Impossible to determine

Commentary

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.

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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?


Commentary

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.”

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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?


Commentary

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.

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

0up

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.

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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

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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 10. Impossible to determine


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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 9. The negative of one of the choices above 10. Impossible to determine



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 traveled, and displacement.

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

9. The negative of one of the choices above 10. Impossible to determine

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 ∆x.

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 traveled, and displacement.

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.

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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 ∆x is the change in position, ∆x 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

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(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 8. Impossible to determine

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 ∆x is the change in position, ∆x 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’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

Purpose: To hone your understanding of “distance traveled.”

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).

Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x


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.

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

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(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

Purpose: To hone the understanding of instantaneous velocity, in particular of its direction.

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.

Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x


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

Purpose: To hone your understanding of (instantaneous) speed.

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.

Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x Amy

Brad

Cate

t

2

1

0

0 1 2 3 4 5time (s)

posi

tion

(m

)

6

�1

�2

x



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 (v

0 = +15.0 m/s) to a value of 8.00 m/s

downwardward (vf = −8.00 m/s) is given by

∆ ∆t

a gf= =

−−

=− − +( )

−v v v0 8 00 15 0

9 80

. .

.

m s m s

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

velocity v0 225= + m s is found from v v202 2= + ( )a y∆ as

∆yg

( ) =− ( )

−( ) =− ( )−(max .

0

2

0 225

2 9 8002 2v m s

m s2 )) = ×2 58 103. m

Thus, the projectile will be at the ∆y = ×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 ∆y t gt= −v0

12

2 and obtaining the largest of the two solutions to the resulting quadratic equation:

6 20 10 2251

29 802 2. .× = ( ) − ( ) m m s m s2t t

or

4 90 225 6 20 10 02 2. . m s m s m2( ) − ( ) + × =t t

The quadratic formula yields

t =− −( ) ± −( ) − ( ) ×225 225 4 4 90 6 20 10

2 2 m s m s m s2. . 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 t t t t= = −( ) = −( ) =∆ ∆v v v v v0 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 ∆ x t t= =v v 2. In the special case where a = = =( )0 00and hence v v , we see that statements (a), (b), (c), and (d) are all cor-rect. However, in the general case a ≠ ≠( )0 0, and hence v 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: y

1 = 4.9 m. The displacement of the pebble at t = 3.0 s is found from

y t3 021

20

1

29 8 3 0 44= + = + ( )( ) =v at 2 . .m s s2 m

The distance fallen in the 2.0 s interval from t = 1.0 s to t = 3.0 s is then

∆y y y= − = − =3 1 44 4 9 39m m 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 ( )∆x x xt tf i

= − .

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 = = − −∆ ∆x t x x t tf 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 ∆y = 30 m. The acceleration may be found using the kinematics equation v v2

02 2= + ( )a y∆ as

ay

=−

( ) =( ) −

( ) =v v2

02 2

2

6 0

2 300 6

∆m s

mm 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 ∆t = = × −1 1 000 1 0 10 3 s s. is given by

∆ ∆x t= ( ) = ( ) ×( ) = × =− −v 8 5 1 0 10 8 5 10 83 3. . . .m s s m 55 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 viR iB= + = −0 0and , respectively. The velocity of either ball when it has a displacement from the launch point of ∆y h= − (where h is the height of the building) is found from v v2 2 2= + ( )i a y∆ as follows:

v v v vR iR Ra y g h gh= − + ( ) = − +( ) + −( ) −( ) = − +2

0

2

022 2 2∆

and

v v v vB iB Ba y g h gh= − + ( ) = − −( ) + −( ) −( ) = − +2

0

2

022 2 2∆

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

v v v vR gh gh= − + = + >02

02

02 2 and v v v vB gh gh= − + = + >02

02

02 2 .

Therefore, v vR 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 ∆y h= − , where h is the height of the tower and upward has been chosen as the positive direction. From v v2

02 2= + ( )a y∆ ,

the speed of the rock just before hitting the ground is found to be

v v v= ± + ( ) = + −( ) −( ) = ( ) +02

02 2

2 2 12 2 9 8a y g h∆ m s m. ss m m s2 40 0 30( )( ) =.

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

∆ ∆t

x= = = × =−

v2

2 10 0 022m

100 m ss s.

2.2 At constant speed, c = ×3 108 m s, the distance light travels in 0.1 s is

∆ ∆x c t= ( ) = ×( )( ) = ×( )3 10 0 1 3 1018 7m s s m

mi

1.6.

009 km

km

10 mmi3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ×12 104

Comparing this to the diameter of the Earth, DE, we fi nd

∆ ∆x

D

x

RE E

= = ××( ) ≈

2

3 0 10

2 6 38 102 4

7

6

.

..

m

m (with RE = Earth’s radius)

2.3 Distances traveled between pairs of cities are

∆ ∆x t1 1 1 80 0 0 500 40 0= km h h kmv ( ) = ( )( ) =. . .

∆ ∆x t2 2 2 1 0 200 20 0= 00 km h h kmv ( ) = ( )( ) =. .

∆ ∆x t3 3 3 40 0 0 750 30 0= km h h kmv ( ) = ( )( ) =. . .

Thus, the total distance traveled is ∆x = + +( ) =40 0 20 0 30 0. . . km 90.0 km, and the elapsed

time is ∆t = + + + =0 0 0 0.500 h .200 h .750 h .250 h 1.70 h.

(a) v = = =∆∆

x

t

90.0 km

1.70 h52.9 km h

(b) ∆x = 90.0 km (see above)



2.4 (a) v = = ⎛⎝⎜

⎞⎠⎟ ×

∆∆

x

t

20 1 1ft

1 yr

m

3.281 ft

yr

3.156 110 sm s7

⎛

⎝⎜⎞

⎠⎟= × −2 10 7

or in particularly windy times,

v = = ⎛⎝⎜

⎞⎠⎟

∆∆

x

t

1 1 100 ft

1 yr

m

3.281 ft

yr

3.156 ××⎛

⎝⎜⎞

⎠⎟= × −

10 sm s7 1 10 6

(b) The time required must have been

∆ ∆t

x= = × ⎛⎝⎜

⎞⎠⎟v

3 10 1609 103 3mi

10 mm yr

m

1 mi

mmm

1 myr

⎛⎝⎜

⎞⎠⎟

= ×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 = =−−

∆∆

x

t

x x

t tf i

f i

(a) v = = −−

=∆∆

x

t

1 0

0

0.0 m

2.00 s5.00 m s

(b) v = = −−

=∆∆

x

t

5 0

0

.00 m

4.00 s1.25 m s

(c) v = = −−

= −∆∆

x

t

5 1.00 m 0.0 m

4.00 s 2.00 s2.50 m s

(d) v = = − −−

= −∆∆

x

t

5 5.00 m .00 m

7.00 s 4.00 s3.33 m s

(e) v = = −−

= −−

=∆∆

x

t

x x

t t2 1

2 1

0 0

8.00 s 00

2.7 (a) Displacement km h min1 h= = ( )( )∆x 85 0 35 0

60 0. .

. minkm km⎛

⎝⎞⎠ + =130 180

(b) The total elapsed time is

∆t = +( )⎛⎝⎜

⎞⎠⎟ +35 0 15 0

12 0. . . min min

h

60.0 min00 2 84 h h= .

so,

v = = =∆∆

x

t

180

2 846

km

h3.4 km h

.


46 Chapter 2

2.8 The average velocity over any time interval is

v = =−−

∆∆

x

t

x x

t tf i

f i

(a) v = = −−

= +∆∆

x

t

4 0

1 0 04 0

.

..

m 0

sm s

(b) v = = − −−

= −∆∆

x

t

2 0

4 0 00 50

.

..

m 0

sm s

(c) v = = −−

= −∆∆

x

t

0 4 0

5 0 1 01 0

.

. ..

m

s sm s

(d) v = = −−

=∆∆

x

t

0

5 0 0

0

s0

.

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 0 501 0 0

1 0 0

4.

.

.ss

s

.0 m

1.0 s=

−−

= ==x x t 4 0. m s

(b) v 22 1 0

2 1 0.0 s.5 s s

.5 s s

6.0 m

1.5 s=

−−

= −x x .

.== − 4 0. m s

(c) v 34 2

2

0.0 s

.0 s .5 s

4.0 s .5 s 1.5 s=

−−

= =x x

0

(d) v 45 4 0

5 4 0.5 s.0 s s

.0 s s

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 tx= ∆

v. Thus, the difference in the times for the

two cars to complete the same 10 mile trip is

∆ ∆ ∆

t t tx x= − = − = −

⎛1 2

1 2

10

55

10

7v vmi

mi h

mi

0 mi h⎝⎝⎜⎞⎠⎟

⎛⎝

⎞⎠ =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 ∆ ∆x t1 1 55 15= ( ) = ( )( )v mi h min .

The faster car pulls ahead of the slower car at a rate of

vrelative mi h 55 mi h 15 mi h= − =70

Thus, the time required for it to get distance ∆x1 ahead is

∆ ∆t

x=

mi h min

mi hrelative

1 55 15

15 0v=

( )( ).

== 55 min

Finally, the distance the faster car has traveled during this time is

∆ ∆x t2 2

1= 70 mi h 55 min

h

60 minv ( ) = ( )( )⎛

⎝⎞⎠ == 6 mi4



2.11 The distance traveled by the space shuttle in one orbit is

Circumference of Orbit 2 Earth s radius= =2π πr ’ + 200 miles

2 3963 200 mi mi

( )= +( ) = ×π 2 61 104.

Thus, the required time is

t = = ×Circumference

average speed

2 61 10

19 8

4. mi

0001 32

mi hh= .

2.12 (a) v11

1 11=

( )( ) = + = +∆∆

x

t

L

tL t

(b) v22

2 22=

( )( ) = − = −∆∆

x

t

L

tL t

(c) vtotaltotal

total

=( )( ) =

( ) + ( )+

=∆∆

∆ ∆x

t

x x

t t1 2

1 2

++ −+

=+

=L L

t t t t1 2 1 2

00

(d) ave speedtotal distance traveled

t.

trip( ) = (∆ )) =( ) + ( )

+= + + −

+=

+total

∆ ∆x x

t t

L L

t t

L

t t1 2

1 2 1 2 1 2

2

2.13 The total time for the trip is t t t= + = +1 122 0. min 0.367 h, where t1 is the time spent

traveling at v1 = 89.5 km h. Thus, the distance traveled is ∆x t t= =v v1 1 , which gives

∆x t t= ( ) = ( ) +( ) =89 5 77 8 0 367 77 81 1. . . .km h km h h km h km( ) +t1 28 5.

or

89 5 77 8 28 51. . . km h km h km−( ) =t

From which, t1 2 44= . h for a total time of t t= + =1 0 367. h 2 80. h .

Therefore, ∆x t= = ( )( ) =v 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 t− = −2 0 120. min s. The speed of the tortoise is vt = 0 100. m s, and the speed of the hare is v vh 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 xt h= + 0 20. m

which becomes

v vt ht t= −( ) +120 0 20s m.

or

0 100 2 0 120 0 20. . . m s m s s m( ) = ( ) −( ) +t t

This gives the time of the race as t = ×1 3 102. s

(b) x tt t= = ( ) ×( ) =v 0.100 m s s m1 3 10 132.


48 Chapter 2

2.15 The maximum allowed time to complete the trip is

t total = total distance

required average speed==

⎛⎝⎜

⎞⎠⎟

=1 600 m

250 km h

1 km h

0.278 m s23.0 s

The time spent in the fi rst half of the trip is

t1

1

= =half distance

v800 m

230 km h

1 km h

0.278 m sss

⎛⎝⎜

⎞⎠⎟

= 12 5.

Thus, the maximum time that can be spent on the second half of the trip is

t t t2 = − =total 1 23 0 12 5 10 5. s . s . s− =

and the required average speed on the second half is

v2

800 m

s76.2 m s

1 km= = =half distance

t2 10 5.

hh

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 t2 2= v and the trailing athlete will travel a distance d t1 1= v . Only when d d d1 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 d1 2 1 2= + = +or v v

giving

v vv v1 2

1 2

t t d td− = =−( )or

(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 td≥ = =−( )

⎡

⎣⎢

⎤

⎦⎥2 2 2

1 2

v vv v

or Dd≥

−v

v v2

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= = −−

=1 00

10 0 0

2 00 05 00.

.

..s

m

sm s

(b) vt= = −( )−( ) = −3 00

5 00 10 0

4 00 2 002 50.

. .

. ..s

m

smm s

(c) vt= = −( )−( ) =4 50

5 00 5 00

5 00 4 000.

. .

. .s

m

s

(d) vt= =− −( )

−( ) =7 50

0 5 00

8 00 7 005 00.

.

. ..s

m

sm 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 = = =∆∆

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 = = =∆∆

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 = = =∆∆

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 = = −−

=∆∆

x

t

6 0

0

8.0 m

4.00 s 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

12

2v with a = 0 for each runner, the x-coordinate of each runner at time t is

x t xA B= − + ( ) =4 0 6 0 3 0. . . mi mi h and mi ++ −( )5 0. mi h t

When the runners meet, x xA B= , giving − + = + −4 0 6 0 3 0 5 0. ( . ) . ( . ) mi mi h mi mi ht t, or ( . . ) . .6 0 5 0 3 0 4 0 mi h mi h mi mi+ = +t . This gives the elapsed time when they meet as

t = =7 00 64

..

mi

11.0 mi h h

At this time, x xA 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 13 0 4 00 1 00 9 00at . . . .m s m s s m s2

(b) At an elapsed time of t = 2 00. s, v = + −( )( ) =13 0 4 00 2 00 5 00. . . .m s m s s m s2 .

(c) When t = 2 50. s, v = + −( )( ) =13 0 4 00 2 50 3 00. . . .m s m s s m s2 .

(d) At t = 4 00. s, v = + −( )( ) = −13 0 4 00 4 00 3 00. . . .m s m s s m s2 .

(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

∆t

During any quarter mile segment, the distance traveled is

∆x = ⎛⎝

⎞⎠ =1 5 280

1 320mi

4

ft

1 mift

(a) During the fi rst quarter mile segment, Secretariat’s average speed was

v1

1 32052 4= =ft

25.2 sft s.

During the second quarter mile segment,

v2

1 32055 0= =ft

24.0 sft s.

For the third quarter mile of the race,

v3

1 32055 5= =ft

23.8 sft s.

For the fourth fi nal quarter mile,

v4

1 32056 9= =ft

23.2 sft s.

and during the fi nal quarter mile,

v5

1 32057 4= =ft

23.0 sft s.

(b) Assuming that v vfinal = 5 and recognizing that v0 0= , the average acceleration for the

entire race was

a =−

= −v vfinal ft s

25.0 57 4 0

total elapsed time

.

22 24.0 23.8 23.2 23.0 sft s2

+ + + +( ) = 0 481.



2.22 From a t= ∆ ∆v , the required time is seen to be

∆ ∆

ta g

g= = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

v 60 0 0

7

1

9 8

.

.

mi h

0 m s

0.2

4447 m s

mi hs

10 391

⎛⎝⎜

⎞⎠⎟

= .

2.23 From a t= ∆ ∆v , we have ∆∆

ta

= = −( ) ⎛⎝⎜

⎞v 60 55

0 60

0 447mi h

m s

m s

1 mi h2.

.⎠⎠⎟

= 3 7. s .

2.24 (a) From t = 0 to t = 5 0. s,

at t

f i

f i

=−−

=− − −( )

−=

v v 8 0 8 0

5 0 00

. .

.

m s m s

s

From to t = 15 s,

a =− −( )−

=8 0 8 0

1 5 01 6

. .

..

m s m s

5 s sm s 2

and from t = 0 to t = 20 s,

a =− −( )

−=

8 0 8 0

2 00 80

. ..

m s m s

0 sm s 2

(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 s 2 , and at t = 18 s, the slope of the

tangent line is 0 .

2.25 (a) at

= = − = ⋅∆∆

v 175 0

2 570 0

mi h

smi h s

..

or

a =⋅

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

70 01609

1

1.

mi

h s

m

mi

h

3 600 s

⎛⎛⎝⎜

⎞⎠⎟

= 31 3. m s2

Alternatively,

as

gg= ⎛

⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

=31 31

9 803 192.

..

m

m s2

(b) If the acceleration is constant, ∆x t at= +v012

2:

∆x = + ⎛⎝

⎞⎠ ( ) =0

1

231 3 2 50 97 82. . .

m

ss m2

or

∆x = ( )⎛⎝

⎞⎠ =97 8

3 281321.

.m

ft

1 mft


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 25 0 0 750 8 50 31 4at . . . .m s m s s2 m s

or

m s eastwardv = 31 4.

(b) When the acceleration is directed westward, a = −0 750. m s2, the fi nal velocity is v v= + = + + −( )( ) = +0 25 0 0 750 8 50 18 6at . . . .m s m s s2 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 ∆t = 3 50. s) is v = +25 0. m s.

(a) The car’s acceleration is

at t

= =−

= − = −∆∆ ∆

v v v0 25 0 40 0

3 504 29

. .

..

m s m s

sm ss or m s westward2 2a = 4 29.

(b) The distance traveled during the 3 50. s time interval is

∆ ∆ ∆x t t= ( ) =+⎛

⎝⎞⎠ = +⎛

⎝⎞⎠v

v vav

m s m s0

2

25 0 40 0

2

. .33 50 114. s m( ) =

2.28 From v v202 2= + ( )a x∆ , we have 10 97 10 0 2 2203 2

. ×( ) = + ( )m s ma so that

ax

=−

( ) =×( ) −( ) =

v v202 3 2

2

10 97 10 0

2 2202 7

∆.

.m s

m44 10

2 74 101

5

5

×

= ×( )⎛⎝⎜

⎞⎠⎟

=

m s

m s9.80 m s

2

22.

g22 79 104. × times !g

2.29 (a) ∆ ∆ ∆x t t= ( ) = +( )v v vav 0 2 becomes 40 02 80

28 500.

..m

m ss=

+⎛⎝

⎞⎠ ( )v

,

which yields v0

2

8 5040 0 2 80 6 61= ( ) − =

.. . .

sm m s m s

(b) at

=−

= − = −v v0 2 80 6 61

8 500 448

∆. .

..

m s m s

sm 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 vf i a x2 2 2= + ( )∆ .

continued on next page



(c) ax

f i=−

( )v v2 2

2 ∆

(d) ax

f i=−

( ) =( ) − ( )

×v v2 2 2 2

2

30 0 20 0

2 2 00 1∆. .

.

m s m s

001 25

2 mm s2

( ) = .

(e) Using a t= ∆ ∆v , we fi nd that ∆∆

ta a

f i= =−

= − =v v v 30 0 20 0

1 258 00

. .

..

m s m s

m s2 ss .

2.31 (a) With v = 120 km h, v v202 2= + ( )a x∆ yields

ax

=−

( ) =( ) −⎡⎣ ⎤⎦

( )v v2

02

2

2

120 0

2 240

0 278

∆

km h

m

. m s

km h1

2⎛⎝⎜

⎞⎠⎟

= 2 32. m s2

(b) The required time is ∆ta

=−

=−( ) ⎛

⎝v v0 120 0

2 32

0 278km h

m s

m s

1 km h2.

.⎜⎜

⎞⎠⎟

= 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:

taspeed

up2

m s

m ss=

−= − =

v v0 20 0

2 010

.

The total time for the trip is t t t ttotal speedup

constantspeed

braking= + + = 10 ss s s s+ + =20 5 0 35. .

(b) The distance traveled during the fi rst 10 s is

∆x t t( ) = = +⎛⎝⎜

⎞⎠⎟speed

upspeedup

speedup

spvv v0

2 eeedup

20 m s 0 s m= +⎛

⎝⎜⎞⎠⎟ ( ) =

210 100

The distance traveled during the next 20 s (with a = 0) is

∆x t( ) = ⋅ = ( )constantspeed

constantspeed

m sv 20 200 400s m( ) =

The distance traveled in the last 5.0 s is

∆x t tf( ) = =+⎛

⎝⎜⎞⎠⎟braking braking braking brv

v v

2 aaking

0 20 m s.0 s m= +⎛

⎝⎞⎠ ( ) =

25 50

The total displacement is then

∆ ∆ ∆ ∆x x x x( ) = ( ) ( ) (+ +total speed

upconstantspeed

)) = + + =braking

m m m 550 m100 400 50

and the average velocity for the entire trip is

vtrip

total

total

m

35 sm s=

( )= =

∆x

t

55016


54 Chapter 2

2.33 (a) at

=−

= − =v v0 24 0 0

2 958 14

∆.

..

m s

sm s

22

(b) From a t= ∆ ∆v , the required time is ∆ta

f i=−

= − =v v 20 0 10 0

8 141 23

. .

..

m s m s

m ss2 .

(c) Yes. For uniform acceleration, the change in velocity ∆v generated in time ∆t is given by ∆ ∆v = ( )a t . From this, it is seen that doubling the length of the time interval ∆t will always

double the change in velocity ∆v. 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

ta

=−

= −−

=v v0 0 100 m s

5.00 m s2 20 0. s

(b) The minimum distance needed to stop is

∆x t t= ⋅ =+⎛

⎝⎞⎠ ⋅ = +⎛

⎝⎞⎠ ( ) =v

v v0

2

0 10020 0 10

m s

2s. 000 1 00m 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 t= =0 0 and 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 S0 0= and v0 30 0S = . m s. Her constant acceleration for t aS≥ = −0 2 00 is m s2. . The initial conditions for the van are x0 0155 5 00V Vv= =m, m s. and its constant acceleration is aV = 0. We then use ∆x x x t at= − = +0 0

12

2v to write an equation for the x coordinate of each vehicle for t ≥ 0. This gives

Sue’s Car:

x t tS − = ( ) + −( )0 30 01

22 00 2. .m s m s or2 xx t tS = ( ) − ( )30 0 1 00 2. .m s m s2

Van:

x t tV − = ( ) + ( )155 5 001

20 2 m m s or . m m sx tV = + ( )155 5 00.

In order for a collision to occur, the two vehicles must be at the same location i.e., x xS 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 xS V= ⇒ 30 0 1 00 155 5 002. . . m s m s m m s2( ) − ( ) = + ( )t t t

or 1 00 25 00 155 02. . m s m s m2( ) − ( ) + =t

Using the quadratic formula yields

t =− −( ) ± −( ) − ( )25 00 25 00 4 1 00 155

2. . . m s m s m s 2 mm

2 m s 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 xS t V tcollision s s

m= = = += =11 4 11 4155 5 00

. .. m s s m( )( ) =11 4 212.



2.36 The velocity at the end of the fi rst interval is

v v= + = + ( ) =0 0 2 77 15 0 41 6at ( . . .m s) s m s

This is also the constant velocity during the second interval and the initial velocity for the third interval.

(a) From ∆x t at= +v021

2, the total displacement is

∆ ∆ ∆ ∆x x x x( ) = ( ) + ( ) + ( )

= + ( )

total

2m s

1 2 3

01

22 77 1. 55 0 4 1 0

4

2. s 1.6 m s 23 s

1

( )⎡⎣⎢

⎤⎦⎥

+ ( )( ) +⎡⎣ ⎤⎦

+ ..6 m s .39 s 9.47 m s .39 s2( )( ) + −( )( )⎡⎣⎢

⎤41

24 2

⎦⎦⎥

or

∆x( ) = + × + = ×total

m m m312 5 11 10 91 2 5 51 103 3. . . mm km= 5 51.

(b) v1

1

1

31220 8=

( )= =

∆x

t

m

15.0 sm s.

v

2

2

2

541 6=

( )= × =

∆x

t

.11 10 m

123 sm s

3

.

v3

3

3

9=( )

= =∆x

t

1.2 m

4.39 s 20 8. m s

and the average velocity for the total trip is

vtotal

total

total

m=( )

= ×+ +

∆x

t

5 51 10

15 0 123 4

3.

. ...

3938 7

( ) =s

m s

2.37 Using the uniformly accelerated motion equation ∆x t at= +v012

2 for the full 40 s interval yields ∆x = ( )( ) + −( )( ) =20 40 1 0 40 01

22m s s m s s2. , which is obviously wrong.

The source of the error is found by computing the time required for the train to come to rest. This time is

ta

=−

= −−

=v v0 0 20

20m s

1.0 m ss2

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 ∆x t at= +v0

12

2 to this interval gives the stopping distance as

∆x = ( )( ) + −( )( ) =20 20 1 0 2012

2m s s m s s2. 200 m


56 Chapter 2

2.38 v v0 0 40 00 447= = ⎛

⎝⎞⎠

⎛⎝

andmi

h

m s

1 mi hf ..

⎜⎜⎞⎠⎟

= 17 9. m s

(a) To fi nd the distance traveled, we use

∆x t tf= ⋅ =+⎛

⎝⎜⎞⎠⎟

⋅ = +⎛⎝

⎞⎠ (v

v v0

2

17 9 0

212 0

..

m ss)) = 107 m

(b) The constant acceleration is

at

f=−

= − =v v0 17 9 0

12 01 49

.

..

m s

sm s2

2.39 At the end of the acceleration period, the velocity is

v v= + = + ( )( ) =0 0 1 5 5 0 7 5ataccel2m s s m s. . .

This is also the initial velocity for the braking period.

(a) After braking, v vf at= + = + −( )( ) =brake2m s 2.0 m s s m7 5 3 0 1 5. . . ss .

(b) The total distance traveled is

∆ ∆ ∆x x x t ttotal accel brake accel= ( ) + ( ) = ⋅( ) + ⋅(v v )) =+⎛

⎝⎞⎠ +

+⎛⎝⎜

⎞⎠⎟brake accel brake

v v v v0

2 2t tf

∆xtotal

m ss

m s m s= +⎛⎝

⎞⎠ ( ) + +7 5 0

25 0

1 5 7 5

2

..

. .⎛⎛⎝

⎞⎠ ( ) =3 0 32. s m

2.40 For the acceleration period, the parameters for the car are: initial velocity = =via 0, acceleration = =a aa 1, elapsed time = ( ) =∆t t

a 1, and final velocity = v fa. For the braking period, the parameters are: initial velocity final vel. of accel. p= =vib eeriod = v fa, acceleration = =a ab 2, and elapsed time = ( ) =∆t t

b 2 .

(a) To determine the velocity of the car just before the brakes are engaged, we apply v vf i a t= + ( )∆ to the acceleration period and fi nd

v v vib fa ia a aa t a t= = + ( ) = +∆ 0 1 1 or vib a t= 1 1

(b) We may use ∆ ∆ ∆x t a ti= ( ) + ( )v 12

2 to determine the distance traveled during the

acceleration period (i.e., before the driver begins to brake). This gives

∆ ∆ ∆x t a t a ta ia a a a( ) = ( ) + ( ) = +v 1

22

1 120

1

2 or ∆x a t

a( ) = 1

2 1 12

(c) The displacement occurring during the braking period is

∆ ∆ ∆x t a t a t t a tb ib b b b( ) = ( ) + ( ) = ( ) +v 1

22

1 1 2 2 221

2

Thus, the total displacement of the car during the two intervals combined is

∆ ∆ ∆x x x a t a t t a ta b( ) = ( ) + ( ) = + +

total

1

2

1

21 12

1 1 2 2 222



2.41 The time the Thunderbird spends slowing down is

∆ ∆ ∆t

x x1

1

1

1

0

2 2 250

0 71 56 99= =

( )+

= ( )+

=v v v

m

m s.. ss

The time required to regain speed after the pit stop is

∆ ∆ ∆t

x x2

2

2

2

0

2 2 350

71 5 09 79= =

( )+

= ( )+

=v v v

m

m s.. ss

Thus, the total elapsed time before the Thunderbird is back up to speed is

∆ ∆ ∆t t t= + + = + + =1 25 00 6 99 5 00 9 79 21 8. . . . .s s s s s

During this time, the Mercedes has traveled (at constant speed) a distance

∆ ∆x tM = ( ) = ( )( ) =v0 71 5 21 8 1 558. .m s s m

and the Thunderbird has fallen behind a distance

d x x xM= − − = − −∆ ∆ ∆1 2 1 558 958m 250 m 350 m = 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 v v= = +∆ ∆x t 0 2 to the entire trip (for which ∆ ∆x d t= + = =4 0 10 0. m, s, and v ) to obtain

d + =

+4 0 0

20. m

10 s

v or v0

4 0= +d . m

5.0 s [1]

Then, applying v v202 2= + ( )a x∆ to the entire trip yields 0 2 4 00

2= + +( )v a d . m .

Substitute for v0 from Equation [1] to fi nd that

04 0

2 4 02

=+( )

+ +( )da d

..

m

25 s m2 and a

d= − + 4 0. m

50 s2 [2]

Finally, apply ∆x t at= +v012

2 to the fi rst 8.0 s of the trip (for which ∆x d= ).

This gives

d a= ( ) + ( )v0128 0 64. s s2 [3]

Substitute Equations [1] and [2] into Equation [3] to obtain

d

d d= +⎛⎝⎜

⎞⎠⎟

( ) + − +⎛⎝⎜

⎞⎠

4 08 0

1

2

4 0..

.m

5.0 ss

m

50 s2 ⎟⎟ ( ) = +64 0 96 3 84s m2 . .d

which yields d = =3 84 96. m 0.04 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 3 0 36 m s s m. in front of the player. At time t > 0, the displacements of the players from their initial positions are

∆x t a tplayer player player .0 m s= ( ) + = +v021

20

1

24 22( ) t 2 [1]

and

∆x t a topponent opponent opponent m= ( ) + =v021

212 ss( ) +t 0 [2]

When the players are side-by-side,

∆ ∆x xplayer opponent m= + 36 [3]

Substituting Equations [1] and [2] into Equation [3] gives

1

24 12 362.0 m s m s m2( ) = ( ) +t t or t t2 6 0 18 0+ −( ) + −( ) =. s s2

Applying the quadratic formula to this result gives

t =− −( ) ± −( ) − ( ) −( )

( )6 0 6 0 4 1 18

2 1

2. . s s s2

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.

(b) ∆x t a tplayer player player .0 m s= ( ) + = +v021

20

1

24 22 s( )( ) =8 2 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

∆x t t= ⋅ = +( )⎡⎣ ⎤⎦v v v0 2 as

tx

=( )+

= ( )+( ) =

2 2 0 400

82 4 16 48 10

0

∆v v

.

. ..

km

km h××( )⎛

⎝⎞⎠ =−10

3 60029 13 h

s

1 hs.

2.45 (a) From v v202 2= + ( )a y∆ with v = 0, we have

∆ya

( ) =−

=− ( )−( ) =

max 2

m s

m s

v v202 2

2

0 25 0

2 9 80

.

. 31 9. m

(b) The time to reach the highest point is

taup 2

m s

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

∆y t at= ( ) +01

22 as t

y

a= ( ) =

−( )−

=2 2 31 9

9 802 55

∆ .

..

m

m s s2

(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 0 9 80 2 55at . .m s s2

− 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 velocity is v = 0. Thus, v v2

02 2= + ( )a y∆ yields

∆ya

( ) =−

=− ( )−( ) =

max .

v v202 2

2

0 100

2 9 80

m s

m s2510 m

(b) The time for the upward fl ight is

ty y

upup o

m

m s=

( )=

( )+

=( )

+=

∆ ∆max max

v v v

2 2 510

100 0110 2. s

For the downward fl ight, ∆ ∆y y a= − ( ) = − = = −max

, .510 0 9 80m, and m s2v . Thus,

∆y t at= +v012

2 gives ty

adown 2

m

m ss=

( ) =−( )

−=

2 2 510

9 8010 2

∆.

.

and the total time of the fl ight is t t ttotal down down s s= + = + =10 2 10 2. . 20 4. s .

2.47 The velocity of the object when it was 30 0. m above the ground can be determined by applying ∆y t at= +v0

12

2 to the last 1 50. s of the fall. This gives

− = ( ) + −⎛⎝

⎞⎠ ( )30 0 1 50

1

29 80 1 500

2. . . .m s

m

ss2v or v0 12 7= − . m s

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 v v2

02 2= + ( )a y∆ as

∆ya

( ) =−

=−( ) −

−( ) = −1

202 2

2

12 7 0

2 9 808 2

v v .

..

m s

m s233 m

The total distance the object drops during the fall is

∆ ∆y y( ) = ( ) + −( ) =total

m m1

30 0 38 2. .

2.48 (a) Consider the rock’s entire upward fl ight, for which v0 7 40= + . m s, v f = 0, a g= − = −9 80. m s2, yi = 1 55. m (taking y = 0 at ground level), and y hf = =max maximum altitude reached by rock . Then applying v vf i a y2 2 2= + ( )∆ to this upward fl ight gives

0 7 40 2 9 80 1 552= ( ) + −( ) −( ). . .max m s m s m2 h

and solving for the maximum altitude of the rock gives

hmax ..

..= +

( )( ) =1 557 40

2 9 804 34

2

mm s

m sm

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 vf i a y2 2 2= + ( )∆ and solve for v f when yf = 3 65. m (starting with vi iy= + =7 40 1 55. .m s at m). This yields

v vf i f ia y y= + −( ) = ( ) + −( )2 22 7 40 2 9 80 3 6. . .m s m s2 55 1 55 3 69m m m s−( ) =. .



60 Chapter 2

(c) A rock thrown downward at a speed of 7 40 7 40. .m s m svi = −( )from the top of the wall undergoes a displacement of ∆y y yf i( ) = − = − = −1 55 3 65 2 10. . . m m m

before reaching

the level of the attacker. Its velocity when it reaches the attacker is

v vf i a y= − + ( ) = − −( ) + −( ) −2 22 7 40 2 9 80 2∆ . . .m s m s2 110 9 79m 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

∆ speed f i( ) = − = − − − =down

m s m s mv v 9 79 7 40 2 39. . . ss

(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

∆ speed f i( ) = − = − =up

m s m s m sv v 3 69 7 40 3 71. . .

Thus, the two rocks undergo the same magnitudedo not 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:

vv v

up

up up 7.40 m s m s

2m s=

+= + =

i f

2

3 695 55

..

and

vv v

down

down down 7.40 m s m s

2=

+= + =

i f

2

9 798 60

.. mm s

Thus, the rock thrown upward spends more time travveling 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 from rest) is given by v v2

02 2= + ( )a y∆ as

v vI a y= − + ( ) = − + −( ) −( ) = −02 2 0 2 9 8 0 40 2 8∆ . . .m s m m s2

If, upon impact, the child’s head undergoes an additional displacement ∆y h= − before coming to rest, the acceleration during the impact can be found from v v2

02 2= + ( )a y∆

to

be a h hI I= − − =( ) ( )0 2 22 2v v . The duration of the impact is found from v v= +0 at as t a hI I= = −∆v 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):

ahI= =

−( )×( ) =−

v2 2

32

2 8

2 2 0 10

.

.

m s

m2 0 103. × m s2

and th

I

= − =− ×( )

−= × =

−−2 2 2 0 10

2 81 4 10

33

v

.

..

m

m ss 1 4. ms

Carpeted Floor ( h = × −1 0 10 2. m ):

ahI= =

−( )×( ) =−

v2 2

22

2 8

2 1 0 10

.

.

m s

m3 9 102. × m s2

and th

I

= − =− ×( )

−= × =

−−2 2 1 0 10

2 87 1 10

23

v

.

..

m

m ss 7 1. ms

2.50 (a) After 2 00. s, the velocity of the mailbag is

v vbag2m s m s s= + = − + −( )( ) = −0 1 50 9 80 2 00 21at . . . ..1 m s

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

∆y t( ) =+⎛

⎝⎞⎠ =

− + −( )⎡⎣⎢

⎤bag

m s m sv v0

2

21 1 1 50

2

. .

⎦⎦⎥( ) = −2 00 22 6. .s m

During this time, the helicopter, moving downward with constant velocity, undergoes a displacement of

∆y t at( ) = + = −( )( ) + = −copter

m s sv021

21 5 2 00 0 3. . ..00 m

The distance separating the package and the helicopter at this time is then

d y yp h

= ( ) − ( ) = − − −( ) = − =∆ ∆ 22 6 3 00 19 6 19. . . . m m m 66 m



62 Chapter 2

(c) Here, v v0 0 1 50( ) = ( ) = +bag copter

m s. and abag2m s= −9 80. while acopter = 0. After 2 00. s,

the velocity of the mailbag is

vbag 2

ms

m

ss

ms= + −⎛

⎝⎞⎠ ( ) = −1 50 9 80 2 00 18 1. . . .

and its speed is

vbag

ms

= 18 1.

In this case, the displacement of the helicopter during the 2.00 s interval is

∆ycopter m s s m= +( )( ) + = +1 50 2 00 0 3 00. . .

Meanwhile, the mailbag has a displacement of

∆y t( ) =+⎛

⎝⎜⎞⎠⎟

= − +⎡⎣bag

bag m s m sv v0

2

18 1 1 50

2

. .⎢⎢

⎤⎦⎥( ) = −2 00 16 6. .s m


d y yp h

= ( ) − ( ) = − − +( ) = − =∆ ∆ 16 6 3 00 19 6 19. . . . m m m 66 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 9 80. . m s m s downward2 2

(b) At its maximum height, the ball comes to rest momentarily and then begins to fall back

downward. Thus, vmaxheight

= 0 .

(c) Consider the relation ∆y t at= +v012

2 with a g= − . When the ball is at the thrower’s hand, the displacement is ∆y = 0, giving 0 0

12

2= −v t gt

This equation has two solutions, t = 0 which corresponds to when the ball was thrown, and t g= 2 0v corresponding to when the ball is caught. Therefore, if the ball is caught at t = 2 00. s, the initial velocity must have been

v0 2

9 80 2 00

2= =

( )( )=gt . .m s s2

9 80. m s

(d) From v v202 2= + ( )a y∆ , with v = 0 at the maximum height,

∆ya

( ) =−

=− ( )−( ) =

max 2

m s

m s

v v202 2

2

0 9 80

2 9 80

.

.4 90. m



2.52 (a) Let t = 0 be the instant the package leaves the helicopter, so the package and the helicopter have a common initial velocity of v vi = − ( )0 choosing upward as positive .

At times t > 0, the velocity of the package (in free-fall with constant acceleration a gp = − )

is given by v v= +i at as v v vp ogt gt= − − = − +( )0 and speed gtp o= = +v v .

(b) After an elapsed time t, the downward displacement of the package from its point of release will be

∆y t a t t gt t gtp i p( ) = + = − − = − +⎛

⎝⎞⎠v v v

1

2

1

2

1

22

02

02

and the downward displacement of the helicopter (moving with constant velocity, or accel-eration ah = 0) from the release point at this time is

∆y t a t t th i h( ) = + = − + = −v v v

1

202

0 0


d y y t gt t gtp h

= ( ) − ( ) = − +⎛⎝

⎞⎠ − −( ) =∆ ∆ v v0

20

21

2

1

2

(c) If the helicopter and package are moving upward at the instant of release, then the com-mon initial velocity is v vi = + 0. The accelerations of the helicopter (moving with constant velocity) and the package (a freely falling object) remain unchanged from the previous case a g ap h= − =( ) and 0 .

In this case, the package speed at time t > 0 is v v v vp i pa t gt gt= + = − = −0 0 .

At this time, the displacements from the release point of the package and the helicopter are given by

∆y t a t t gtp i p( ) = + = −v v

1

2

1

22

02 and ∆y t a t t t

h i h( ) = + = + = +v v v1

202

0 0

The distance separating the package and helicopter at time t is now given by

d y y t gt t gtp h

= ( ) − ( ) = − − =∆ ∆ v v02

021

2

1

2 (the same as earlier!)

2.53 (a) After its engines stop, the rocket is a freely falling body. It continues upward, slowing under the infl uence of gravity until it comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls.

(b) When it reaches a height of 150 m, the speed of the rocket is

v v= + ( ) = ( ) + ( )( ) =02 2

2 50 0 2 2 00 150a y∆ . .m s m s m2 555 7. m s

After the engines stop, the rocket continues moving upward with an initial velocity of v0 55 7= . m s and acceleration a g= − = −9 80. m s2. When the rocket reaches maximum height, v = 0. The displacement of the rocket above the point where the engines stopped (that is, above the 150-m level) is

∆ya

=−

=− ( )−( ) =

v v202 2

2

0 55 7158

. m s

2 9.80 m sm

2

The maximum height above ground that the rocket reaches is then given by

hmax m m m= + =150 158 308 .



64 Chapter 2

(c) The total time of the upward motion of the rocket is the sum of two intervals. The fi rst is the time for the rocket to go from v0 50 0= . m s at the ground to a velocity of v = 55 7. m s at an altitude of 150 m. This time is given by

ty y

11

1

1

0 2

2 150

50 0=

( )=

( )+( ) = ( )

+( )∆ ∆v v v

m

55.7 . m ss= 2 84.

The second interval is the time to rise 158 m starting with v0 55 7= . m s and ending with v = 0. This time is

ty y

22

2

2

0 2

2 1585=

( )=

( )+( ) = ( )

+=

∆ ∆v v v

m

0 55.7 m s..67 s

The total time of the upward fl ight is then t t tup s= + = +( ) =1 2 2 84 5 67. . 8 51. s

(d) The time for the rocket to fall 308 m back to the ground, with v0 0= and acceleration a = −g = −9.80 m/s2, is found from ∆y t at= +v0

12

2 as

ty

adown 2

m

m ss=

( ) =−( )

−=

2 2 308

9 807 93

∆.

.

so the total time of the fl ight is t t tflight up down s s= + = +( ) =8 51 7 93 16 4. . . .

2.54 (a) The camera falls 50 m with a free-fall acceleration, starting with v0 10= − m s. Its velocity when it reaches the ground is

v v= + ( ) = −( ) + −( ) −( ) =02 2 22 10 2 9 80 50a y∆ m s m s m. −− 33 m s

The time to reach the ground is given by

ta

=−

=− − −( )

−=

v v0 33 10m s m s

9.80 m s2 2 3. s

(b) This velocity was found to be v = − 33 m s in part (a) above.

2.55 During the 0.600 s required for the rig to pass completely onto the bridge, the front bumper of the tractor moves a distance equal to the length of the rig at constant velocity of v = 100 km h. Therefore, the length of the rig is

L trig

km

h

m s

km h= =

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥v 100

0 278

10 6

.. 000 16 7s m( ) = .

While some part of the rig is on the bridge, the front bumper moves a distance

∆x L L= + =bridge rig 400 16 7 m m+ .

With a constant velocity of v = 100 km h, the time for this to occur is

tL L

=+

= +bridge rig m m

100 km h

1 km h

v400 16 7

0 2

.

. 778 m s=

⎛⎝⎜

⎞⎠⎟

15 0. s



2.56 (a) From ∆x t at= +v012

2, we have 100 30 0 3 5012

2m m s m s2= ( ) + −( ). .t t . This reduces to 3 50 60 0 200 02. ( . ( )t t+ − + =s) s2 , and the quadratic formula gives

t =− −( ) ± −( ) − ( )( )60 0 60 0 4 3 50 200

2 3 5

2. . .

.

s s s2

00( ) The desired time is the smaller solution of t = 4 53. s . The larger solution of t = 12 6. s is

the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration.

(b) The velocity of the boat when it fi rst reaches the buoy is

v v= + = −( )( ) =0 30 0 4 53at . .m s+ 3.50 m s s2 14 1. m s

2.57 (a) The acceleration of the bullet is

ax

=−

( ) = ( ) − ( )v v202

2 0 100∆300 m/s 400 m/s

2

2 2

. mm( ) = − ×3.50 10 m s5 2

(b) The time of contact with the board is

ta

=−

= −( )− ×

=v v0 300 400 m s

3.50 10 m s5 2 2 86 10 4. × − s

2.58 We assume that the bullet begins to slow just as the front end touches the fi rst surface of the board, and that it reaches its exit velocity just as the front end emerges from the opposite face of the board.

(a) The acceleration is

ax

=−

( ) =( ) − ( )v vexit

2 m s 20 m s02 2 2

2

280 4

2 0 100∆ . mm s2

( ) = − ×4 90 105.

(b) The average velocity as the front of the bullet passes through the board is

vv v

=+

= + =exit m s 20 m sm s0

2

280 4

2350

and the total time of contact with the board is the time for the front of the bullet to pass through plus the additional time for the trailing end to emerge (at speed vexit),

tx L

=( )

+ = +∆

board bullet

exit

m

350 m sv v0 100 0 0. . 2200

3 57 10 4m

280 m ss= × −.

(c) From v v202 2= + ( )a x∆ , with v = 0, gives the required thickness is

∆xa

=−

=− ( )

− ×( ) =v v2

02 2

52

0 420

2 4 90 100 1

m s

m s2.. 880 18 0m cm= .


66 Chapter 2

2.59 (a) The keys have acceleration a g= − = −9.80 m s2 from the release point until they are caught 1.50 s later. Thus, ∆y t at= +v0

12

2 gives

v0

2 22 4 00 9 80 1 50 2

= − =+( ) − −( )( )∆y at

t

. . .m m s s2

11 5010 0

..

sm s= +

or

v0 = 10 0. m s upward

(b) The velocity of the keys just before the catch was

v v= + = + −( )( ) = −0 10 0 9 80 1 50 4 68at . . . .m s m s s m2 ss

or

v = 4 68. m s downward

2.60 (a) The keys, moving freely under the infl uence of gravity a g= −( ), undergo a vertical displacement of ∆y h= + in time t. We use ∆y t ati= +v 1

22 to fi nd the initial velocity as

h t g ti= + −( )v1

22

giving

vi

h gt

t

h

t

gt= + = +2 2

2

(b) The velocity of the keys just before they were caught at time t( ) is given by v v= +i at as

v = +⎛⎝

⎞⎠ + −( ) = + − = −h

t

gtg t

h

t

gtgt

h

t

gt

2 2 2

2.61 (a) From v v202 2= + ( )a y∆ , the insect’s velocity after straightening its legs is

v v= + ( ) = + ( ) ×( ) =−02 32 0 2 4 000 2 0 10 4 0a y∆ m s m m s2 . .

and the time to reach this velocity is

ta

=−

= − = × =−v v02

34 0 01 0 10 1 0

.. .

m s

4 000 m ss ms

(b) The upward displacement of the insect between when its feet leave the ground and it comes to rest momentarily at maximum altitude is

∆ya g

=−

=−−( ) =

− ( )−( ) =

v v v202

02 2

2

0

2

4 0

2 9 8

.

.

m s

m s200 82. m



2.62 The distance required to stop the car after the brakes are applied is

∆xa

( ) =−

=−

⎛⎝

stop

mi

h

ft s

mi hv v202

2

0 35 01 47

1.

.⎜⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−( ) =

2

2 9 00147

. ft sft

2

Thus, if the deer is not to be hit, the maximum distance the car can travel before the brakes are applied is given by

∆ ∆x x( ) = − ( ) = − =before stop

ft ft ft200 200 147 53..0 ft

Before the brakes are applied, the constant speed of the car is 35.0 mi/h. Thus, the time required for it to travel 53.0 ft, and hence the maximum allowed reaction time, is

tx

r( ) =( )

=max

before ft

35.0mi

h

ft

∆v0

53 0

1 47

.

. ss

1 mi h

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= 1 03. s

2.63 The falling ball moves a distance of 15 m −( )h before they meet, where h is the height above the ground where they meet. Apply ∆y t at= +v0

12

2, with a g= − , to obtain

− −( ) = −15 0 12

2m h gt or h gt= −15 12

2m [1]

Applying ∆y t at= +v012

2 to the rising ball gives

h t gt= ( ) −25 12

2m s [2]

Combining equations [1] and [2] gives

251

215

1

22 2 m s m( ) − = −t gt gt

or

t = =15

250 60

m

m s s.

2.64 The constant speed the student has maintained for the fi rst 10 minutes, and hence her initial speed for the fi nal 500 yard dash, is

v010 1 0 500 5 280 1 500

= = − =−∆

∆x

t

. mi yards

10 min

ft ftt

600 s

m

3.281 ftm s

( ) ⎛⎝⎜

⎞⎠⎟

=11 9.

With an initial speed of v0 1 9= . m s and constant acceleration of a = 0 15. m s2, the maximum distance the student can travel in the remaining 2.0 min (120 s) of her allotted time is

∆x t a t2 0 012

2 1 9 1201

20. max max .( ) = + = ⎛

⎝⎞⎠ ( ) +v

ms s .. .15 120 1 3 10

2 3m

ss m2

⎛⎝

⎞⎠ ( ) = ×

or

∆x2 031 3 10

3 281 1. max

..( ) = ×( )⎛

⎝⎜⎞⎠⎟

mft

1 m

yard

3 fttyards

⎛⎝⎜

⎞⎠⎟

= ×1 4 103.

Since ∆x2 0. max( ) is considerably greater than the 500 yards she must still run, she can

easily meet the requirment of running 1.0 miles in 12 minutes.


68 Chapter 2

2.65 We solve Part (b) of this problem fi rst.

(b) When the either ball reaches the ground, its displacement from the balcony is ∆y = −19 6. m (taking upward as positive). The initial velocities of the two balls were v01 14 7= − . m s and v02 14 7= + . m s, so v0

2 has the value of 14 72

. m s( ) for either ball. Also, a g= − for each ball, giving the downward velocity of either ball when it reaches the ground as

v veitherball

m s= − + ( ) = − ( ) + −02 2

2 14 5 2 9 80a y∆ . . mm s m m s2( ) −( ) = −19 6 24 5. .

(a) The time for either ball to reach the ground (and hence achieve the velocity computed above) is given by

ta g

=−

=− −

−=

+v v

v veitherball m s m s0

0 024 5 24 5

9

. .

..80 m s2

where v0 is the initial velocity of the particular ball of interest.

For ball 1, v0 14 7= − . m s, giving

t1

24 5 14 7

9 801 00= − =. .

..

m s m s

m s s2

For ball 2, v0 14 7= + . m s, and

t2

24 5 14 7

9 804 00= + =. .

..

m s m s

m s s2

The difference in the time of fl ight for the two balls is seen to be

∆t t t= − = −( ) =2 1 4 00 1 00 3 00. . .s s

(c) At t = 0 800. s, the displacement of each ball from the balcony (at height h above ground) is

y h t gt1 0121

214 7 0 800 4 90− = − = −( )( ) −v . . .m s s m s22 s( )( )0 800 2.

y h t gt2 0221

214 7 0 800 4 90− = − = +( )( ) −v . . .m s s m s22 s( )( )0 800 2.

These give the altitudes of the two balls at t = 0 800. s as y h1 14 9= − . m and y h2 8 62= + . m. Therefore the distance separating the two balls at this time is

d y y h h= − = + − −( )2 1 8 62 14 9 23 5. . . m m = m

2.66 (a) While in the air, both balls have acceleration a a g1 2= = − (where upward is taken as positive). Ball 1 (thrown downward) has initial velocity v v01 0= − , while ball 2 (thrown upward) has initial velocity v v02 0= + . Taking y = 0 at ground level, the initial y-coordinate of each ball is y y h01 02= = + . Applying ∆y y y t ati i= − = +v 1

22 to each ball gives their

y-coordinates at time t as

Ball 1: y h t g t1 021

2− = − + −( )v or y h t gt1 0

21

2= − −v

Ball 2: y h t g t2 021

2− = + + −( )v or y h t gt2 0

21

2= + −v




(b) At ground level, y = 0. Thus, we equate each of the equations found above to zero and use the quadratic formula to solve for the times when each ball reaches the ground. This gives the following:

Ball 1: 01

22 2 00 1 1

212

0 1= − − → + ( ) + −( ) =h t gt t t hv vg

so tg h

g g g10 0

2

0 0

22 2 4 2

2=

− ± ( ) − ( ) −( )= − ±

⎛⎝⎜

⎞⎠⎟

+v v v v 22h

g

Using only the positive solution gives

tg g

h

g10 0

22= − +

⎛⎝⎜

⎞⎠⎟

+v v

Ball 2: 01

22 20 2 2

222

0 2= + − → + −( ) + −( ) =h t gt t t hv vg 00

and

tg h

g g g20 0

2

0 02 2 4 2

2=

− −( ) ± −( ) − ( ) −( )= + ±

⎛⎝⎜

⎞v v v v⎠⎠⎟

+2

2h

g

Again, using only the positive solution

tg g

h

g20 0

22= +

⎛⎝⎜

⎞⎠⎟

+v v

Thus, the difference in the times of fl ight of the two balls is

∆t t tg g

h

g g g= − = +

⎛⎝⎜

⎞⎠⎟

+ − − +⎛⎝⎜

⎞⎠⎟2 1

0 0

2

0 0

22v v v v

++⎛

⎝⎜⎜

⎞

⎠⎟⎟

=2 2 0h

g g

v

(c) Realizing that the balls are going downward v <( )0 as they near the ground, we use v vf i a y2 2 2= + ( )∆ with ∆y h= − to fi nd the velocity of each ball just before it strikes the ground:

Ball 1: v v v v1 12

1 0

2

022 2 2f i a h g h g= − + −( ) = − −( ) + −( ) −( ) = − + hh

Ball 2: v v v v2 22

2 0

2

022 2 2f i a h g h g= − + −( ) = − +( ) + −( ) −( ) = − + hh

(d) While both balls are still in the air, the distance separating them is

d y y h t gt h t gt o= − = + −⎛⎝

⎞⎠ − − −⎛

⎝⎞⎠ =2 1 0

20

21

2

1

22v v v tt


70 Chapter 2

2.67 (a) The fi rst ball is dropped from rest v01 0=( ) from the height h of the window. Thus, v vf a y2

02 2= + ( )∆ gives the speed of this ball as it reaches the ground (and hence the initial

velocity of the second ball) as v vf a y g h gh= + ( ) = + −( ) −( ) =012

1 12 0 2 2∆ . When ball 2 is thrown upward at the same time that ball 1 is dropped, their y-coordinates at time t during the fl ights are given by y y t ato− = +v0

12

2

as

Ball 1: y h t g t112

20− = ( ) + −( ) or y h gt112

2= −

Ball 2: y gh t g t212

20 2− = ( ) + −( ) or y gh t gt212

22= ( ) −

When the two balls pass, y y1 2= , or

h gt gh t gt− = ( ) −12

2 12

22

giving

th

gh

h

g= = = ( ) =

2 2

28 71 21

..

m

2 9.80 m s s

2

(b) When the balls meet,

th

g=

2

and

y h gh

gh

h h1

21

2 2 4

3

4= −

⎛

⎝⎜⎞

⎠⎟= − = .

Thus, the distance below the window where this event occurs is

d h y hh h= − = − = = =1

3

4 4

28 77 18

..

m

4 m

2.68 We do not know either the initial velocity nor the fi nal velocity (that is, velocity just before impact) for the truck. What we do know is that the truck skids 62.4 m in 4.20 s while accelerating at −5 60. m s2.

We have v v= +0 at and ∆x t t= ⋅ = + ⋅v v v[( ) ]0 2 . Applied to the motion of the truck, these yield

v v v v− = − = −( )( ) = −0 0 5 60 4 20 2at or m s s2. . 33 5. m s [1]

and

v v+ =( ) = ( )

=0

2 2 62 4

4 2029 7

∆x

t

.

..

m

sm s [2]

Adding equations [1] and [2] gives the velocity just before impact as

2 23 5 29 7v = − +( ). . m s, or v = 3 10. m s

2.69 When released from rest v0 0=( ), the bill falls freely with a downward acceleration due to gravity a g= − = −( )9 80. m s2 . Thus, the magnitude of its downward displacement during David’s 0.2 s reaction time will be

∆y t at= + = + −( )( ) = =v02 21

20

1

29 80 0 2 0 2. . .m s s m2 220 cm

This is over twice the distance from the center of the bill to its top edge ≈( )8 cm , so

David will be unsuccessful .



2.70 (a) The velocity with which the fi rst stone hits the water is

v v1 012

2

2 2 00 2 9 80= − + ( ) = − −⎛⎝

⎞⎠ + −⎛

⎝a y∆ . .ms

m

s2⎞⎞⎠ −( ) = −50 0 31 4. .m

ms

The time for this stone to hit the water is

ta1

1 0131 4 2 00

=−

=− − −( )⎡⎣ ⎤⎦

−v v . .m s m s

9.80 m s2 == 3 00. s

(b) Since they hit simultaneously, the second stone which is released 1.00 s later will hit the water after an fl ight time of 2.00 s. Thus,

v0222

2

22 50 0 9 80 2 00

= − =− − −( )( )∆y at

t

. . .m m s s2 22

2 0015 2

..

sm s= −

(c) From part (a), the fi nal velocity of the fi rst stone is v1 31 4= − . m s .

The fi nal velocity of the second stone is

v v2 02 2 15 2 9 80 2 00 34= + = − + −( )( ) = −at . . .m s m s s2 ..8 m s

2.71 (a) The sled’s displacement, ∆x1, while accelerating at a1 40= + ft s2 for time t1 is

∆x t a t t1 112 1 1

2120 20= ( ) + = ( )ft s2 or ∆x t1 1

220= ( )ft s2 [1]

At the end of time t1, the sled had achieved a velocity of

v v= + = + ( )0 1 1 10 40a t tft s2 or v = ( )40 1ft s2 t [2]

The displacement of the sled while moving at constant velocity v for time t2 is

∆x t t t2 2 1 240= = ( )⎡⎣ ⎤⎦v ft s2 or ∆x t t2 1 240= ( )ft s2 [3]

It is known that ∆ ∆x x1 2 17 500+ = ft, and substitutions from Equations [1] and [3] give

20 40 17 50012

1 2ft s ft s ft2 2( ) + ( ) =t t t or t t t12

1 22 875+ = s2 [4]

Also, it is known that

t t1 2 90+ = s [5]

Solving Equations [4] and [5] simultaneously yields

t t t12

1 12 90 875+ −( ) = s s2 or t t12

1180 875 0+ −( ) + = s s2

The quadratic formula then gives

t1

2180 180 4 1 875

2 1=

− −( ) ± −( ) − ( )( )( )

s s s2

with solutions t t1 25 00 90 5 0 85= = − =( ). .s and s s s or t t1 2175 85= = −( )s and s .

Since it is necessary that t2 0> , the valid solutions are t t1 25 0 85= =. s and s .

(b) From Equation [2] above, v = ( ) = ( )( ) =40 40 5 0 2001ft s ft s s ft s2 2t . .



72 Chapter 2

(c) The displacement ∆x3 of the sled as it comes to rest (with acceleration a3 20= − ft s2) is

∆xa3

2

3

20

2

200

2 201 000= − =

− ( )−( ) =v ft s

ft sft

2

Thus, the total displacement for the trip (measured from the starting point) is

∆ ∆ ∆ ∆x x x xtotal ft ft= +( ) + = + =1 2 3 17 500 1 000 18 500 fft

(d) The time required to come to rest from velocity v (with acceleration a3) is

ta3

3

0 200

2010= − = −

−=v ft s

ft ss2

so the duration of the entire trip is t t t ttotal s s s s= + + = + + =1 2 3 5 0 85 10 100. .

2.72 (a) From ∆y t at= +v021

2 with v0 0= , we have

ty

a= ( ) =

−( )−

=2 2 23

9 80

∆ m

m s2.2 2. s

(b) The fi nal velocity is v = + −( )( ) =0 29.80 m s m s .2 s2 2 − 21 m s .

(c) The time it takes for the sound of the impact to reach the spectator is

ty

soundsound

m

340 m ss= = = × −∆

v23

6 8 10 2.

so the total elapsed time is t total s s= + × ≈−2 2 6 8 10 2. . 2 3. s .

2.73 (a) Since the sound has constant velocity, the distance it traveled is

∆x t= = ( )( ) =vsound ft s s1100 5 0. 5 5 103. × ft

(b) The plane travels this distance in a time of 5 0 10. s s 15 s+ = , so its velocity must be

vplane

ft

15 s= = × =∆x

t

5 5 103.3 7 102. × ft s

(c) The time the light took to reach the observer was

tx

lightlight

8

ft

3.00 10 m s

m s

3.= = ×

×∆

v5 5 10 13.

2281 ft ss

⎛⎝⎜

⎞⎠⎟

= × −5 6 10 6.

During this time the plane would only travell a distance of 0.002 ft .



2.74 The distance the glider moves during the time ∆td is given by ∆ ∆ ∆x t a td d= ( ) + ( )� = v012

2,

where v0 is the glider’s velocity when the fl ag fi rst enters the photogate and a is the glider’s acceleration. Thus, the average velocity is

vv

vdd

d d

ddt

t a t

ta t= =

( ) + ( )= + ( )�

∆∆ ∆

∆∆0

12

2

0

1

2

(a) The glider’s velocity when it is halfway through the photogate in space i.e., when ∆x =( )� 2 is found from v v2

02 2= + ( )a x∆ as

v v v v v v1 02

02

02

022 2= + ( ) = + = + ( )⎡⎣ ⎤⎦ = +a a a t ad d� � ∆ vvd dt∆( )

Note that this is not equal to unlessvd a = 0 , in which case v v v1 0= =d .

(b) The speed v2 when the glider is halfway through the photogate in time (i.e., when the elapsed time is t td2 2= ∆ ) is given by v v= +0 at as

v v v v= + = + ( ) = + ( )0 2 0 021

2at a t a td d∆ ∆

which is equal to vd for all possible values of v0 and a.

2.75 The time required for the stunt man to fall 3.00 m, starting from rest, is found from ∆y t at= +v0

12

2 as

− = + −( )3 00 01

29 80 2. . m m s2 t so t = ( ) =

2 3 00

9 800 782

.

..

m

m s s2

(a) With the horse moving with constant velocity of 10.0 m s, the horizontal distance is

∆x t= = ( )( ) =vhorse m s s m10 0 0 782 7 82. . .

(b) The required time is t = 0 782. s as calculated above.


Solucionario Fundamentos de Física 9na edición Capitulo 2

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Transcript of Solucionario Fundamentos de Física 9na edición Capitulo 2