216

17. v=

dx

dt

=

10–3t

2

a=

dv

dt

= —6t

= —18 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 car’s

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 –6 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 Galileo’s 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 it’s 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 aren’t constant, but

it’s hard to know how they’d really look. We just

idealize them as constant accelerations. Similarly, the

hand’s force on the ball during the time ab is probably

not constant, but we can draw it that way, since we

don’t know how to draw it more realistically. Since our

acceleration graph consists of constant-acceleration

Solutions to Selected Problems