3.00m
20.0
o
A
–
B
3.83m
θ
3.53m
10.0m/s
θ
Introductory Physics I
PHYS 2211160
Exam #1
Chapters 14
Name_____________________________
Show all your work on these sheets, and make sure to box your answers!
For problems 1 – 4 the position of an object is described by the equation
x
(t) = [3m + (4m/s) t + (2m/s
3
) t
3
]
x
.
1.
(10 points)
What is the displacement vector of the object from t
1
= 2s to t
2
= 4s ?
We have that
∆
x
=
x
2
–
x
1
, and with
x
2
=
x
(4s) = [3m + (4m/s) (4s) + (2m/s
3
) (4s)
3
]
x
= 147m
x
, and
x
1
=
x
(2s) = [3m + (4m/s) (2s) + (2m/s
3
) (2s)
3
]
x
= 27m
x
, so
∆
x
= 147m
x

27m
x
=
∆
x
= 120m
x
.
2.
(10 points)
What is the average velocity vector of the object from t
1
= 2s to t
2
= 4s ?
We have that
v
av
=
∆
x
/
∆
t = 120m
x
/ (4s – 2s) =
v
av
= (60m/s)
x
.
3.
(10 points)
What is the velocity vector as a function of time for the object?
We have that
v
(t) = d
x
/ dt = (d / dt) [3m + (4m/s) t + (2m/s
3
) t
3
]
x
=
[0 + (4m/s) + (6m/s
3
) t
2
]
x
=
v
(t) = [(4m/s) + (6m/s
3
) t
2
]
x
.
4.
(10 points)
What is the acceleration vector as a function of time for the object?
We have that
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 Spring '11
 lomant
 Acceleration, Velocity, 4s, 2S, velocity vector

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