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001
10.0 points
A ball is dropped from rest at point
O.
It
passes a window with height 3
.
8 m in time
interval
t
AB
= 0
.
02 s.
Assume:
Down is the positive
y
direction.
Use
g
= 9
.
8 m
/
s
2
.
O
A
B
3
.
8 m
x
y
bardbl
vectorv
bardbl
=
v
is the speed of the ball.
Identify the correct pair of equations, which
enable us to solve for speed
v
B
.
1.
v
A
−
v
B
=
gt
AB
,
v
A
+
v
B
2
=
h
t
AB
2.
v
A
−
v
B
=
gt
AB
, v
A
+
v
B
=
h
t
AB
3.
v
B
−
v
A
=
g t
AB
,
v
A
+
v
B
2
=
h
t
AB
cor
rect
4.
v
B
−
v
A
=
gt
AB
, v
A
+
v
B
=
h
t
AB
Explanation:
Let :
h
= 3
.
8 m
,
t
AB
= 0
.
02 s
,
and
g
= 9
.
8 m
/
s
2
.
Basic Concept
The change in speed over
time is acceleration by definition
g
=
v
B
−
v
A
t
AB
.
Note:
The change in speed is the difference
between the final and the initial speed over a
time interval.
O
A
B
h
t
AB
t
y
Solution
v
B
is the final speed over the
time needed to pass the window, so it appears
first in the expression. The average speed for
the ball can be expressed in terms of a dis
placement over time or as an average of the
final and initial speed (provided the accelera
tion is constant); thus
v
Avg
=
v
A
+
v
B
2
=
h
t
AB
.
002
(part 1 of 3) 10.0 points
The position of a softball tossed vertically
upward is described by the equation
y
=
c
1
t
−
c
2
t
2
,
where
y
is in meters,
t
in seconds,
c
1
=
7
.
02 m
/
s, and
c
2
= 3
.
57 m
/
s
2
.
Find the ball’s initial speed
v
0
at
t
0
= 0 s.
Correct answer: 7
.
02 m
/
s.
Explanation:
Basic Concepts:
v
=
dx
dt
a
=
dv
dt
=
d
2
x
dt
2
Solution:
The velocity is simply the derivative of
y
with respect to
t
:
v
=
dy
dt
= 7
.
02 m
/
s
−
2(3
.
57 m
/
s
2
)
t,
which at
t
= 0 is
v
0
= 7
.
02 m
/
s
.
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 Fall '08
 Turner
 Acceleration, Work, Velocity, Correct Answer, m/s

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