Solutions Manual - College Physics: Explore and Apply, 3rd Edition
by Eugenia Etkina, Planinsic, Rutberg, Heuvelen
!"#$% &' (&)*%)*+
Chapter 2 Kinematics: Motion in One Dimension
Chapter 3 Newtonian Mechanics
Chapter 4 Applying Newton’s Laws
Chapter 5 Circular Motion
Chapter 6 Impulse and Linear Momentum
Chapter 7 Work and Energy
Chapter 8 Extended Bodies at Rest
Chapter 9 Rotational Motion
Chapter 10 Vibrational Motion
Chapter 11 Mechanical Waves
Chapter 12 Gases
Chapter 13 Static Fluids
Chapter 14 Fluids in Motion
Chapter 15 First Law of Thermodynamics
Chapter 16 Second Law of Thermodynamics
Chapter 17 Electric Charge, Force, and Energy
Chapter 18 The Electric Field
Chapter 19 DC Circuits
Chapter 20 Magnetism
Chapter 21 Electromagnetic Induction
Chapter 22 Reflection and Refraction
Chapter 23 Mirrors and Lenses
Chapter 24 Wave Optics
Chapter 25 Electromagnetic Waves
Chapter 26 Special Relativity
Chapter 27 Quantum Optics
Chapter 28 Atomic Physics
Chapter 29 Nuclear Physics
Chapter 30 Composition of the Universe
,Chapter 2
Multiple choice questions
1. (c) A rolling ball is an example of a physical phenomenon. A point-like object is a simplified model of an
object. Acceleration is a physical quantity for describing motion, while free fall is a model of a process.
2. (b) Average speed, path length, and clock reading are all scalar quantities. Displacement and acceleration are
examples of vector quantities.
3. (b) A time interval is the difference between two instants in time. Both statements, (2) The lesson lasted for 45
minutes and (4) An astronaut orbited Earth in 4 hours, are examples of time intervals.
4. (a) The student should have said “The distance between my dorm and the lecture hall is 1 km.” There is no
indication of the direction (which is needed for indicating displacement). The path length depends on the path
followed and that is also not indicated.
5. (b) With x = +12 m − (4 m/s)t + (1 m/s )t
2 2
the corresponding velocity and acceleration are
v x (t) = −4 m/s + (2 m/s )t
2
and a x (t) = +2 m/s
2
. Therefore, we see that the object is always accelerating with
a x = +2.0 m/s
2
. The speed of the object first decreases, reaches zero at t = 2.0 s , and then increases beyond that.
So (b) is not true.
6. (a) The motion of the car is described by graph (a). The positive flat part of the velocity curve corresponds to
the car moving at a constant positive velocity. The car then slows down (as indicated by the negative slope of the
velocity curve), comes to a stop (v = 0 when the velocity curve goes through the x-axis), and then moves in the
opposite direction (indicated by a negative velocity) with the same acceleration.
7. (a), (c) At t1 , the curve of position versus time is flat, so the position of the dot is not changing in time here.
Therefore, its velocity is zero and choice (a) is true. The dot has nonzero speed where the slope of the position
versus time graph is nonzero (such as at t2 ), so the maximum speed is not at t1 and choice (b) is false. However,
by measuring the displacement on the vertical x-axis, we find that the dot has moved from point C in the +x -
direction to point D at time t1 , so choice (c) is correct. At t2 , the dot has moved in the −x -direction with respect
to point D, but it has not moved far enough to reach point A. Thus, choice (d) is incorrect. Finally, the slope of
the position versus time curve is negative at t2 , so the velocity is negative at this time and choice (d) is incorrect.
8. (a) Assume the downward direction is positive. At the instant the second marble is released, the first marble
has traveled a certain distance and acquired a positive velocity. The two marbles then continue on with the same
positive acceleration (due to gravity). However, because the first marble had already acquired a positive velocity,
it will always be traveling faster than the second marble, so the distance between them will increase with time
and choice (a) is correct. Mathematically, the positions of the two marbles are and
1 2
y 1 (t) = g(t + Δt)
2
y 2 (t) =
1
2
gt
2
, where Δt is the time interval between dropping the first marble and the second. The distance
between them is given by , which shows clearly that increases with t.
1 2
Δy = y 1 − y 2 (t) = (gΔt)t + g(Δt) Δy
2
This result is also revealed by a position versus time graph or a motion diagram. The left figure below shows a
position versus time graph of y 1 (t) and y 2 (t) for Δt = 1 s . The blue shaded area corresponds to the difference
Δy(t) and shows that Δy(t) increases with time t. The right figure below shows a motion diagram for the first 5 s
of free fall for the two marbles, again with Δt = 1 s . The difference in position Δy again increases in with time t,
as shown explicitly by comparing Δy(t = 1 s) with Δy(t = 4 s) .
, Extended Figure Description for Chapter 2 Multiple choice questions Problem 8 Figure 1
9. (b) The position of the car can be written as x(t) = x 0 + v 0x t +
1
2
ax t
2
, where x 0 = +20 m . Since the car is
traveling west (in the −x -direction), v 0x = −12 m/s . With v x (t) = v 0x + a x t , the acceleration ax must be positive
to bring the velocity of the car to zero at the stoplight.
10. (c) The velocity-versus-time graph in (c) describes the motion of the car with v x (t) = v 0x + a x t ,
v 0x = −12 m/s at t = 0 , ax > 0 (i.e., positive slope of velocity versus time curve), and v(t) = 0 at t > 0 .
11. (d) Azra needs to count the number n of poles she passes in a given time interval Δt, then divide the total
distance traveled ( = nd ) by the time interval Δt . This gives v = nd/Δt , which is choice (d) with Δt = 10 s .
12. (c) At the moment the sandbag is released, it has the same upward velocity as the hot air balloon, according
to the ground observer 2. Therefore, he sees the sandbag going up first then coming down. On the other hand,
observer 1 in the hot air balloon sees the sandbag undergo free fall from the instant it is released.
2 2
v (5.0 m/s)
13. (b) The height of the tree is h = =
2
= 1.3 m . The correct choice is (b).
2g 2(9.8 m/s )
14. (c) Whether you drop a ball or throw it down, the acceleration of the ball is due to the gravitational force
exerted by Earth, so it remains the same in both situations: a y = −g = −9.8 m/s
2
(where the upward direction is
+y ). So statement (c) is incorrect.
15. (c), (e) An object that undergoes zero displacement has zero total area between its velocity-versus-time graph
line and the time axis, which is true only for choices (c) and (e).
16. (a) The total flight time is given by , where
t = 2v 0 /g v0 is the initial speed. The fact that t is linear in v0 means
that, if it takes twice as much time for the second ball to come back, the initial speed of the second ball must be
twice that of the first ball.
, Conceptual questions
17. See graphical addition of vectors below. Note that, when adding or subtracting parallel vectors, the vectors
are offset for clarity.
Extended Figure Description for Chapter 2 Conceptual questions Problem 17 Figure 1
18. All graphs except (a) could be correct depending on the initial position and the initial direction of motion.
Graphs (b) and (e) assume that at time zero, Peter is at the origin traveling in the positive direction. Graph (c) is
the same only at a positive position at time zero. Graphs (d) and (f) assume that at time zero, Peter is at a
positive position moving in the negative direction.
19. One scenario is as follows: As the light turns green, the car starts to accelerate from rest. Upon reaching the
appropriate speed, the driver stops accelerating and the car moves at a constant speed. Upon seeing a red light
some distance ahead, the driver starts to brake. With constant deceleration, the car comes to a complete stop at
the light. The velocity-versus-time and acceleration-versus-time graphs are shown below.
Extended Figure Description for Chapter 2 Conceptual questions Problem 19 Figure 1
Another possibility is that the car starts from rest with constant acceleration. Seeing another traffic light a short
distance ahead, the driver abruptly applies the brake and slows the car to a complete stop. The velocity-versus-
time and acceleration-versus-time graphs are shown below.
Extended Figure Description for Chapter 2 Conceptual questions Problem 19 Figure 2
by Eugenia Etkina, Planinsic, Rutberg, Heuvelen
!"#$% &' (&)*%)*+
Chapter 2 Kinematics: Motion in One Dimension
Chapter 3 Newtonian Mechanics
Chapter 4 Applying Newton’s Laws
Chapter 5 Circular Motion
Chapter 6 Impulse and Linear Momentum
Chapter 7 Work and Energy
Chapter 8 Extended Bodies at Rest
Chapter 9 Rotational Motion
Chapter 10 Vibrational Motion
Chapter 11 Mechanical Waves
Chapter 12 Gases
Chapter 13 Static Fluids
Chapter 14 Fluids in Motion
Chapter 15 First Law of Thermodynamics
Chapter 16 Second Law of Thermodynamics
Chapter 17 Electric Charge, Force, and Energy
Chapter 18 The Electric Field
Chapter 19 DC Circuits
Chapter 20 Magnetism
Chapter 21 Electromagnetic Induction
Chapter 22 Reflection and Refraction
Chapter 23 Mirrors and Lenses
Chapter 24 Wave Optics
Chapter 25 Electromagnetic Waves
Chapter 26 Special Relativity
Chapter 27 Quantum Optics
Chapter 28 Atomic Physics
Chapter 29 Nuclear Physics
Chapter 30 Composition of the Universe
,Chapter 2
Multiple choice questions
1. (c) A rolling ball is an example of a physical phenomenon. A point-like object is a simplified model of an
object. Acceleration is a physical quantity for describing motion, while free fall is a model of a process.
2. (b) Average speed, path length, and clock reading are all scalar quantities. Displacement and acceleration are
examples of vector quantities.
3. (b) A time interval is the difference between two instants in time. Both statements, (2) The lesson lasted for 45
minutes and (4) An astronaut orbited Earth in 4 hours, are examples of time intervals.
4. (a) The student should have said “The distance between my dorm and the lecture hall is 1 km.” There is no
indication of the direction (which is needed for indicating displacement). The path length depends on the path
followed and that is also not indicated.
5. (b) With x = +12 m − (4 m/s)t + (1 m/s )t
2 2
the corresponding velocity and acceleration are
v x (t) = −4 m/s + (2 m/s )t
2
and a x (t) = +2 m/s
2
. Therefore, we see that the object is always accelerating with
a x = +2.0 m/s
2
. The speed of the object first decreases, reaches zero at t = 2.0 s , and then increases beyond that.
So (b) is not true.
6. (a) The motion of the car is described by graph (a). The positive flat part of the velocity curve corresponds to
the car moving at a constant positive velocity. The car then slows down (as indicated by the negative slope of the
velocity curve), comes to a stop (v = 0 when the velocity curve goes through the x-axis), and then moves in the
opposite direction (indicated by a negative velocity) with the same acceleration.
7. (a), (c) At t1 , the curve of position versus time is flat, so the position of the dot is not changing in time here.
Therefore, its velocity is zero and choice (a) is true. The dot has nonzero speed where the slope of the position
versus time graph is nonzero (such as at t2 ), so the maximum speed is not at t1 and choice (b) is false. However,
by measuring the displacement on the vertical x-axis, we find that the dot has moved from point C in the +x -
direction to point D at time t1 , so choice (c) is correct. At t2 , the dot has moved in the −x -direction with respect
to point D, but it has not moved far enough to reach point A. Thus, choice (d) is incorrect. Finally, the slope of
the position versus time curve is negative at t2 , so the velocity is negative at this time and choice (d) is incorrect.
8. (a) Assume the downward direction is positive. At the instant the second marble is released, the first marble
has traveled a certain distance and acquired a positive velocity. The two marbles then continue on with the same
positive acceleration (due to gravity). However, because the first marble had already acquired a positive velocity,
it will always be traveling faster than the second marble, so the distance between them will increase with time
and choice (a) is correct. Mathematically, the positions of the two marbles are and
1 2
y 1 (t) = g(t + Δt)
2
y 2 (t) =
1
2
gt
2
, where Δt is the time interval between dropping the first marble and the second. The distance
between them is given by , which shows clearly that increases with t.
1 2
Δy = y 1 − y 2 (t) = (gΔt)t + g(Δt) Δy
2
This result is also revealed by a position versus time graph or a motion diagram. The left figure below shows a
position versus time graph of y 1 (t) and y 2 (t) for Δt = 1 s . The blue shaded area corresponds to the difference
Δy(t) and shows that Δy(t) increases with time t. The right figure below shows a motion diagram for the first 5 s
of free fall for the two marbles, again with Δt = 1 s . The difference in position Δy again increases in with time t,
as shown explicitly by comparing Δy(t = 1 s) with Δy(t = 4 s) .
, Extended Figure Description for Chapter 2 Multiple choice questions Problem 8 Figure 1
9. (b) The position of the car can be written as x(t) = x 0 + v 0x t +
1
2
ax t
2
, where x 0 = +20 m . Since the car is
traveling west (in the −x -direction), v 0x = −12 m/s . With v x (t) = v 0x + a x t , the acceleration ax must be positive
to bring the velocity of the car to zero at the stoplight.
10. (c) The velocity-versus-time graph in (c) describes the motion of the car with v x (t) = v 0x + a x t ,
v 0x = −12 m/s at t = 0 , ax > 0 (i.e., positive slope of velocity versus time curve), and v(t) = 0 at t > 0 .
11. (d) Azra needs to count the number n of poles she passes in a given time interval Δt, then divide the total
distance traveled ( = nd ) by the time interval Δt . This gives v = nd/Δt , which is choice (d) with Δt = 10 s .
12. (c) At the moment the sandbag is released, it has the same upward velocity as the hot air balloon, according
to the ground observer 2. Therefore, he sees the sandbag going up first then coming down. On the other hand,
observer 1 in the hot air balloon sees the sandbag undergo free fall from the instant it is released.
2 2
v (5.0 m/s)
13. (b) The height of the tree is h = =
2
= 1.3 m . The correct choice is (b).
2g 2(9.8 m/s )
14. (c) Whether you drop a ball or throw it down, the acceleration of the ball is due to the gravitational force
exerted by Earth, so it remains the same in both situations: a y = −g = −9.8 m/s
2
(where the upward direction is
+y ). So statement (c) is incorrect.
15. (c), (e) An object that undergoes zero displacement has zero total area between its velocity-versus-time graph
line and the time axis, which is true only for choices (c) and (e).
16. (a) The total flight time is given by , where
t = 2v 0 /g v0 is the initial speed. The fact that t is linear in v0 means
that, if it takes twice as much time for the second ball to come back, the initial speed of the second ball must be
twice that of the first ball.
, Conceptual questions
17. See graphical addition of vectors below. Note that, when adding or subtracting parallel vectors, the vectors
are offset for clarity.
Extended Figure Description for Chapter 2 Conceptual questions Problem 17 Figure 1
18. All graphs except (a) could be correct depending on the initial position and the initial direction of motion.
Graphs (b) and (e) assume that at time zero, Peter is at the origin traveling in the positive direction. Graph (c) is
the same only at a positive position at time zero. Graphs (d) and (f) assume that at time zero, Peter is at a
positive position moving in the negative direction.
19. One scenario is as follows: As the light turns green, the car starts to accelerate from rest. Upon reaching the
appropriate speed, the driver stops accelerating and the car moves at a constant speed. Upon seeing a red light
some distance ahead, the driver starts to brake. With constant deceleration, the car comes to a complete stop at
the light. The velocity-versus-time and acceleration-versus-time graphs are shown below.
Extended Figure Description for Chapter 2 Conceptual questions Problem 19 Figure 1
Another possibility is that the car starts from rest with constant acceleration. Seeing another traffic light a short
distance ahead, the driver abruptly applies the brake and slows the car to a complete stop. The velocity-versus-
time and acceleration-versus-time graphs are shown below.
Extended Figure Description for Chapter 2 Conceptual questions Problem 19 Figure 2
