Patel, Kinal – Homework 2 – Due: Sep 7 2007, 7:00 pm – Inst: D Weathers
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 2) 10 points
The velocity oF a particle moving along the
x
axis varies in time according to the expression
v
(
t
) = (
α

β t
2
)
where
α
= 44
.
2 m
/
s,
β
= 5
.
09 m
/
s
3
, and
t
is
in seconds.
±ind the average acceleration in the time
interval From
t
= 0 to 2
.
68 s.
Correct answer:

13
.
6412 m
/
s
2
.
Explanation:
The velocities at
t
i
= 0 and
t
f
= 2
.
68 s are
Found by substituting these values into the
expression given For the velocity:
v
i
= (
α

β t
2
i
)
= (44
.
2 m
/
s

(5
.
09 m
/
s
3
) (0 s)
2
)
= 44
.
2 m
/
s
,
v
f
= (
α

β t
2
f
)
= (44
.
2 m
/
s

(5
.
09 m
/
s
3
) (2
.
68 s)
2
)
= 7
.
64158 m
/
s
.
ThereFore,
the average acceleration in the
specifed time interval Δ
t
=
t
f

t
i
= 2
.
68 s is
¯
a
=
v
f

v
i
t
f

t
i
=
7
.
64158 m
/
s

44
.
2 m
/
s
2
.
68 s

0 s
=

13
.
6412 m
/
s
2
.
002
(part 2 oF 2) 10 points
Determine the acceleration oF the particle at
t
f
= 2
.
68 s.
Correct answer:

27
.
2824 m
/
s
2
.
Explanation:
We can use the rules oF the di²erential calcu
lus to fnd the velocity From the displacement,
at
t
= 2
.
68 s , the acceleration is
a
(
t
) =
dv
dt
=
d
dt
(44
.
2 m
/
s

5
.
09 m
/
s
3
t
2
)
=

(2) (5
.
09 m
/
s
3
)
t
=

(2) (5
.
09 m
/
s
3
) (2
.
68 s)
=

27
.
2824 m
/
s
2
.
keywords:
003
(part 1 oF 2) 10 points
An electron in the cathode ray tube oF a tele
vision set enters a region where it acceler
ates uniFormly From a speed oF 34600 m
/
s to
a speed oF 4
.
67
×
10
6
m
/
s in a distance oF
2
.
95 cm.
What is its acceleration?
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 Fall '07
 Weathers
 Acceleration, Work, m/s, patel

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