, CHAPTER 2
Quick Quizzes
1. (a) 200 yd (b) 0 (c) 0
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 acceleration 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-time graph (a) has a constant slope, indicating a constant acceleration, which
is represented by acceleration-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-time graph
that best indicates this is (d).
Graph (c) depicts an object that first has a velocity that increases at a constant rate, which
means its 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).
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.
19
, C H A P T E R 2
6. (c). 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 first jumper will always be moving with a higher velocity than the second.
Thus, in a given time interval, the first jumper covers more distance than the second.
Thus, the separation distance between them increases. At any given instant of time, the
velocities of the jumpers are definitely different, because one had a head start. In a time
interval after this instant, however, 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.
20
, C H A P T E R 2
Problem Solutions
2.1 Distances traveled are
∆x 1 = v 1 ( ∆t1 ) = ( 80.0 km h ) ( 0.500 h ) = 40.0 km
∆x 2 = v 2 ( ∆t 2 ) = (100 km h ) ( 0.200 h ) = 20.0 km
∆x 3 = v 3 ( ∆t 3 ) = ( 40.0 km h ) ( 0.750 h ) = 30.0 km
Thus, the total distance traveled is ∆x = ( 40.0 + 20.0 + 30.0 ) km = 90.0 km , and the elapsed
time is ∆t = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h .
∆x 90.0 km
(a) v= = = 52.9 km h
∆t 1.70 h
(b) ∆x = 90.0 km (see above)
∆x 20 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞ −7
2.2 (a) v= = ⎜ ⎟⎜ ⎟ = 2 × 10 m s ,
∆t 1 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
or in particularly windy times
∆x 100 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞ −6
v= = ⎜ ⎟⎜ ⎟ = 1 × 10 m s
∆t 1 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
(b) The time required must have been
∆x 3 × 103 mi ⎛ 1609 m ⎞ ⎛ 103 mm ⎞
∆t = = ⎟ = 5 × 10 yr .
8
⎜ ⎟⎜
v 10 mm yr ⎝ 1 mi ⎠ ⎝ 1 m ⎠
2.3 (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 .
∆x 20 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞
2.4 (a) v= = ⎟ = 5 × 10 m s
-11
⎜ ⎟⎜
∆t 4000 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
21
Quick Quizzes
1. (a) 200 yd (b) 0 (c) 0
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 acceleration 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-time graph (a) has a constant slope, indicating a constant acceleration, which
is represented by acceleration-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-time graph
that best indicates this is (d).
Graph (c) depicts an object that first has a velocity that increases at a constant rate, which
means its 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).
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.
19
, C H A P T E R 2
6. (c). 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 first jumper will always be moving with a higher velocity than the second.
Thus, in a given time interval, the first jumper covers more distance than the second.
Thus, the separation distance between them increases. At any given instant of time, the
velocities of the jumpers are definitely different, because one had a head start. In a time
interval after this instant, however, 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.
20
, C H A P T E R 2
Problem Solutions
2.1 Distances traveled are
∆x 1 = v 1 ( ∆t1 ) = ( 80.0 km h ) ( 0.500 h ) = 40.0 km
∆x 2 = v 2 ( ∆t 2 ) = (100 km h ) ( 0.200 h ) = 20.0 km
∆x 3 = v 3 ( ∆t 3 ) = ( 40.0 km h ) ( 0.750 h ) = 30.0 km
Thus, the total distance traveled is ∆x = ( 40.0 + 20.0 + 30.0 ) km = 90.0 km , and the elapsed
time is ∆t = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h .
∆x 90.0 km
(a) v= = = 52.9 km h
∆t 1.70 h
(b) ∆x = 90.0 km (see above)
∆x 20 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞ −7
2.2 (a) v= = ⎜ ⎟⎜ ⎟ = 2 × 10 m s ,
∆t 1 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
or in particularly windy times
∆x 100 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞ −6
v= = ⎜ ⎟⎜ ⎟ = 1 × 10 m s
∆t 1 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
(b) The time required must have been
∆x 3 × 103 mi ⎛ 1609 m ⎞ ⎛ 103 mm ⎞
∆t = = ⎟ = 5 × 10 yr .
8
⎜ ⎟⎜
v 10 mm yr ⎝ 1 mi ⎠ ⎝ 1 m ⎠
2.3 (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 .
∆x 20 ft ⎛ 1 m ⎞ ⎛ 1 yr ⎞
2.4 (a) v= = ⎟ = 5 × 10 m s
-11
⎜ ⎟⎜
∆t 4000 yr ⎝ 3.281 ft ⎠⎝ 3.156 × 10 s ⎠
7
21
