Solutions:
Exam A
1.
(1)
(D)
m/s
4
x
s
4
=
m
(2)
(E) The initial velocity is (–3
m/s
) (term linear in
t
)
(3)
(B)
4
5
3 2
4 (2)
63
x
m
= 
p
+
=
(4)
(C)
4
(5)
( )
(
)
3
3
( )
3
4
0
3
4 4
3 16
dx
d
d t
d t
v t
t
t
dt
dt
dt
dt
=
=

+
=  +
=  +
(5)
(A) The position vs time graph curves upwards with increasing time.
2.
(6)
(C)
(7)
(D)
0.7
1.5
cg
v
=
m/s
(8)
(A)
2
2
(
)
(
)
cb
bg
cg
v
v
v
=
+
(9)
(D)
tan
bg
cg
v
v
θ
=
(10)
(B)
2
2
2
2
0.7
0.467
/ ;
( ) ( ) (0.467) (0.6) 0.760
1.5
cg
cb
cg
bg
v m s v v v m s
= = = + = + =
1
1
0
0.6
tan ( ) tan ( ) 52.1 ;
0.467
bg
cg
v
v


= = =
3.
(11)
(B)


cos
W
mg
=
P
(for the angle drawn)
(12)
(C)
sin
fr
F
N
mg
μ
=
=
(for the angle drawn)
(13)
(D)
net
F
ma
=
(14)
(A)
cos
sin
mg
mg
ma

=
(15)
(E)
0
0
cos
9.81 cos60
1.8
cos
sin
0.365
sin
9.81 sin60
g
a
mg
mg
ma
g



=#
=
=
=
4.
(16)
(B) The acceleration of gravity
acting on the satellite is
2
)
/(
h
R
GM
+
(17)
(C) The centripetal force is
2
/(
)
GMm R
h
+
(18)
(D)
2
/
mv
GMm r
=
1
cg
v
r
v
r
bg
v
r
C)
0
60
θ=
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(C)
2
v
r f
π
=
(20)
(B)
2
2
/
/
/
mv
r
GmM r
v
GM r
=
=
=
11
24
6
6
6.67 10
6.0 10 /(6.4 10
5.5 10 )
5799
/
m s

=
p
+
=
1
6
6
5
5799
1
24 3600
2
7.76 10
6.70
2
2
.
10
5.5 10 )
v
s
v
r f
f
day
r
s
day


=↵
=
=
=
=
♦(6 4♠
+
5.
(21)
(C) average velocity = 0, since the change in position is 0, but the distance moved is
nonzero.
(22)
(B)
both arrive after the same time on the ground
(23)
(A)
2
2
4
4
4
(0.301 0.1)
1.19
/
g
m
g
m
m s
2
2
=
∆ =
∆
=
=
6.
(24)
(C)
Head To Tail method gives for the vectors
B
A
/
r
and
shown:
(25)
(B)
t = trailor, p = passenger, g = ground, call east positive:
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 Spring '08
 STEPHENS
 Velocity, Trigraph, DT DT DT

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