homework 02 – FONTENOT, BRIAN – Due: Jan 28 2008, 4:00 am
1
Question 1, chap 2, sect 5.
part 1 of 3
10 points
In deep space (no gravity), the bolt (arrow)
of a crossbow accelerates at 129 m
/
s
2
and
attains a speed of 125 m
/
s when it leaves the
bow.
For how long is it accelerated?
Correct answer:
0
.
968992
s (tolerance
±
1
%).
Explanation:
v
=
v
o
+
at
=
at
since the initial speed
v
o
= 0, so
t
=
v
a
=
125 m
/
s
129 m
/
s
2
= 0
.
968992 s
Question 2, chap 2, sect 5.
part 2 of 3
10 points
What speed will the bolt have attained 4
.
5 s
after leaving the crossbow?
Correct answer: 125 m
/
s (tolerance
±
1 %).
Explanation:
It has left the crossbow, so
a
= 0 and the
velocity of the bolt after 4
.
5 s will remain
125 m
/
s.
Question 3, chap 2, sect 5.
part 3 of 3
10 points
How far will the bolt have traveled during
the 4
.
5 s?
Correct answer: 562
.
5 m (tolerance
±
1 %).
Explanation:
s
=
s
o
+
v
o
t
+
1
2
at
2
=
vt
2
since
s
o
= 0,
v
o
=
v
,
a
= 0, and
t
=
t
2
, so
s
= (125 m
/
s)(4
.
5 s)
= 562
.
5 m
Question 4, chap 2, sect 5.
part 1 of 2
10 points
A speed
boat moving at 30
.
2 m
/
s ap
proaches
a
nowake
buoy
marker
75
.
4
m
ahead. The pilot slows the boat with a con
stant acceleration of

4
.
2 m
/
s
2
by reducing
the throttle.
How long does it take the boat to reach the
buoy?
Correct answer: 3
.
21578 s (tolerance
±
1 %).
Explanation:
Given :
v
i
= 30
.
2 m
/
s
Δ
x
= 75
.
4 m
a
=

4
.
2 m
/
s
2
Under the deceleration,
Δ
x
=
v
i
t
+
1
2
a t
2
a t
2
+ 2
v
i
t

2 Δ
x
= 0
so that
t
=

2
v
i
±
radicalbig
4
v
i
2
+ 8
a
(Δ
x
)
2
a
=

2(30
.
2)
±
radicalBig
4(30
.
2)
2
+ 8(

4
.
2)(75
.
4)
2 (

4
.
2)
≈
11
.
1652
,
3
.
21578
.
The larger solution of 11
.
1652 is the second
time the boat would pass the buoy (mov
ing backwards), assuming it could maintain a
constant deceleration after it had stopped.
Question 5, chap 2, sect 5.
part 2 of 2
10 points
What is the velocity of the boat when it
reaches the buoy?
Correct answer: 16
.
6937 m
/
s (tolerance
±
1
%).
Explanation:
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homework 02 – FONTENOT, BRIAN – Due: Jan 28 2008, 4:00 am
2
v
f
=
v
i
+
a t
= 30
.
2 m
/
s +
(

4
.
2 m
/
s
2
)
(3
.
21578 s)
=
16
.
6937 m
/
s
.
Question 6, chap 2, sect 6.
part 1 of 1
10 points
A
ball is
thrown
straight up
from
the
ground with an initial velocity of 37
.
2 m
/
s;
at the same instant, a ball is dropped from
the roof of a building 20
.
8 m high.
After how long will the balls be at the same
height?
Correct answer: 0
.
55914 s (tolerance
±
1 %).
Explanation:
Basic Concepts:
In the case of each ball
the expression
d
=
d
0
+
v
0
t

1
2
gt
2
is adequate to describe the situation.
Solution:
Since the falling ball has no initial
velocity we obtain
d
1
=
h

1
2
gt
2
For the second ball,
r
0
>
0 and
d
0
= 0, so
d
2
=
v
0
t

1
2
gt
2
.
To find the time when the two balls are at
equal height from the ground, we need to
equate the two equations that we have devel
oped:
h

1
2
gt
2
=
v
0
t

1
2
gt
2
which simplifies to
h
=
v
0
t
and we solve for t.
Question 7, chap 2, sect 5.
part 1 of 1
10 points
A baby sitter pushing a stroller starts from
rest and accelerates uniformly at a rate of
0.705 m/s
2
.
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 Spring '08
 Turner
 Physics, Acceleration, Gravity, Work, Velocity, Correct Answer, Fontenot, BRIAN

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