216
17. v=
dx
dt
=
10Ė3t
2
a=
dv
dt
= ó6t
= ó18 m/s
2
18. (a) Solving
.
x=
1
2
at
2
for a, we find a=2
.
x/t
2
=5.51
m/s
2
. (b) v=
2a
.
x
=66.6 m/s. (c) The actual carís
final velocity is less than that of the idealized con-
stant-acceleration car. If the real car and the idealized
car covered the quarter mile in the same time but the
real car was moving more slowly at the end than the
idealized one, the real car must have been going
faster than the idealized car at the beginning of the
race. The real car apparently has a greater accelera-
tion at the beginning, and less acceleration at the
end. This make sense, because every car has some
maximum speed, which is the speed beyond which it
cannot accelerate.
19. Since the lines are at intervals of one m/s and
one second, each box represents one meter. From
t=0 to t=2 s, the area under the curve represents a
positive
.
x of 6 m. (The triangle has half the area of
the 2x6 rectangle it fits inside.) After t=2 s, the area
above the curve represents negative
.
x. To get Ė6 m
worth of area, we need to go out to t=6 s, at which
point the triangle under the axis has a width of 4 s
and a height of 3 m/s, for an area of 6 m (half of 3x4).
20. (a) We choose a coordinate system with positive
pointing to the right. Some people might expect that
the ball would slow down once it was on the more
gentle ramp. This may be true if there is significant
friction, but Galileoís experiments with inclined planes
showed that when friction is negligible, a ball rolling
on a ramp has constant acceleration, not constant
speed. The speed stops increasing as quickly once
the ball is on the more gentle slope, but it still keeps
on increasing. The a-t graph can be drawn by in-
specting the slope of the v-t graph.
v
t
a
t
(b) The ball will roll back down, so the second half of
the motion is the same as in part a. In the first (rising)
half of the motion, the velocity is negative, since the
motion is in the opposite direction compared to the
positive x axis. The acceleration is again found by
inspecting the slope of the v-t graph.
v
t
a
t
21. This is a case where itís probably easiest to draw
the acceleration graph first. While the ball is in the air
(bc, de, etc.), the only force acting on it is gravity, so
it must have the same, constant acceleration during
each hop. Choosing a coordinate system where the
positive x axis points up, this becomes a negative
acceleration (force in the opposite direction com-
pared to the axis). During the short times between
hops when the ball is in contact with the ground (cd,
ef, etc.), it experiences a large acceleration, which
turns around its velocity very rapidly. These short
positive accelerations probably arenít constant, but
itís hard to know how theyíd really look. We just
idealize them as constant accelerations. Similarly, the
handís force on the ball during the time ab is probably
not constant, but we can draw it that way, since we
donít know how to draw it more realistically. Since our
acceleration graph consists of constant-acceleration
Solutions to Selected Problems
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