le (txl65) – Homework 3 – Spurlock – (11881)
1
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printout
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have
6
questions.
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before answering.
001
10.0points
Initially (at time
t
= 0) a particle is mov
ing vertically at 7
.
4 m
/
s and horizontally at
0 m
/
s. Its horizontal acceleration is 1
.
3 m
/
s
2
.
At what time will the particle be traveling
at 48
◦
with respect to the horizontal?
The
acceleration due to gravity is 9
.
8 m
/
s
2
.
Correct answer: 0
.
658141 s.
Explanation:
Let :
v
y
0
= 7
.
4 m
/
s
,
g
= 9
.
8 m
/
s
2
,
v
x
0
= 0
,
and
a
= 1
.
3 m
/
s
2
,
and
θ
= 48
◦
.
v
x
t
v
y
t
v
t
48
◦
The vertical velocity is
v
y
t
=
v
y
0

g t
and the horizontal velocity is
v
x
t
=
v
y
0
+
a t
=
a t .
The vertical component is the opposite side
and the horizontal component the adjacent
side to the angle, so
tan
θ
=
v
y
t
v
x
t
=
v
y
0

g t
a t
a t
tan
θ
=
v
y
0

g t
a t
tan
θ
+
g t
=
v
y
0
t
=
v
y
0
a
tan
θ
+
g
=
7
.
4 m
/
s
(1
.
3 m
/
s
2
) tan(48
◦
) + 9
.
8 m
/
s
2
=
0
.
658141 s
.
002
10.0points
Particle 1 is moving on the
x
axis with an
acceleration of 3
.
37 m
/
s
2
in the positive
x

direction.
Particle 2 is moving on the
y

axis with an acceleration of 5
.
16 m
/
s
2
in the
negative
y
direction.
Both particles were at
rest at the origin at
t
= 0 s.
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 Spring '11
 sdsd
 Acceleration, Velocity, m/s, vy

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