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PHY 201, HW
Conceptual
8. (a) At the maximum height, the ball is momentarily at rest. (That is, it has zero velocity.) The acceleration
remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even
though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the
downward direction.
(b) The acceleration of the ball remains constant in magnitude and direction throughout the ball’s free flight,
from the instant it leaves the hand until the instant just before it strikes the ground. The acceleration is
directed downward and has a magnitude equal to the free-fall acceleration g.
9. a): East – speeding up.
b): East – slowing down
c): East – constant speed
d): West – speeding up
e): West – slowing down
f): West – constant speed
g): Speed up East from rest position.
h): Speed up west from rest position.
Problems
2.8 The average velocity over any time interval is
f i
f i
x^ x^ x
t t t
v
(a)
4.0 m 0 4.0 m s 1.0 s 0
x
t
v
(b)
2 .0 m 0 0.50 m s 4.0 s 0
x
t
v
(c)
0 4.0 m 1.0 m s 5.0 s 1.0 s
x
t
v
(d)
5.0 s 0
x
t
v
2.17 The instantaneous velocity at any time is the slope of the x vs. t graph at
that time. We compute this slope by using two points on a straight
segment of the curve, one point on each side of the point of interest.
(a) 1.00 s
10.0 m 0 5.00 m s 2 .00 s 0
vt
(b) 3.00 s
5.00 10.0 m 2 .50 m s 4.00 2 .00 s
v t
(c) 4.50 s
5.00 5.00 m 0 5.00 4.00 s
v t
(d) 7.50 s
0 5.00 m 5.00 m s 8.00 7.00 s
v t
2.25 (a)
175 mi h 0 70.0 mi h s 2.5 s
a t
v
or
mi a 70. h
1609 m
s 1 mi
1 h (^2) 31.3 m s 3 600 s
Alternatively,
as 21.1 m/s.
(b) The displacement of the mailbag after 2.00 s is
0 bag
21.1 m s 1.50 m s 2 .00 s 22 .6 m 2 2
y t
v v
During this time, the helicopter, moving downward with constant velocity, undergoes a displacement of
2 copter^0
1.5 m s 2 .00 s 0 3.00 m 2
y vt at
The distance separating the package and the helicopter at this time is then
22.6 m 3.00 m 19.6 m 19.6 m p h d y y
(c) Here, v 0^ (^) bag v (^0) copter 1.50 m s and 2 a bag 9.80 m s while a copter = 0. After 2.00 s, the
velocity of the mailbag is
bag (^2)
m m m 1.50 9.80 2.00 s 18. s (^) s s
v
and its speed is
bag
m
s
v
In this case, the displacement of the helicopter during the 2.00 s interval is
y copter 1.50 m s 2.00 s 0 3.00 m
Meanwhile, the mailbag has a displacement of
bag 0 bag
18.1 m s 1.50 m s 2 .00 s 16.6 m 2 2
y t
v v
The distance separating the package and the helicopter at this time is then
16.6 m 3.00 m 19.6 m 19.6 m p h d y y
