College physics 11th edition serway
solutions manual
,College Physics 11th Edition Serway Solutions Manual
Full Download: http://testbanklive.com/download/college-physics-11th-edition-serway-solutions-manual/
Topic 2
Topic 2
Motion in One Dimension
QUICK QUIZZES
2.1 (a) 200 yd (b) 0 (c) 0
2.2 (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 car4the acceleration is negative and goes to zero as the car
comes to rest.
' 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
,2.3 The velocity-vs-time graph (a) has a constant slope, indicating a constant
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 behaviour is (d).
Graph (c) depicts an object which first has a velocity that increases at a
constant rate, which means that the object9s 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).
2.4 Choice (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.
2.5 (a) The blue graph of Figure 2.14b best shows the puck9s 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 puck9s 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.
2.6 Choice (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.
2.7 Choice (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.
2.8 Choices (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, and 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.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
solutions manual
,College Physics 11th Edition Serway Solutions Manual
Full Download: http://testbanklive.com/download/college-physics-11th-edition-serway-solutions-manual/
Topic 2
Topic 2
Motion in One Dimension
QUICK QUIZZES
2.1 (a) 200 yd (b) 0 (c) 0
2.2 (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 car4the acceleration is negative and goes to zero as the car
comes to rest.
' 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
,2.3 The velocity-vs-time graph (a) has a constant slope, indicating a constant
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 behaviour is (d).
Graph (c) depicts an object which first has a velocity that increases at a
constant rate, which means that the object9s 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).
2.4 Choice (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.
2.5 (a) The blue graph of Figure 2.14b best shows the puck9s 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 puck9s 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.
2.6 Choice (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.
2.7 Choice (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.
2.8 Choices (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, and 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.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
