homework 03 – ASH, BEN – Due: Feb 6 2008, 11:00 pm
1
Question 1, chap 3, sect 99.
part 1 of 2
10 points
A ship cruises forward at
v
s
= 6 m
/
s rel
ative to the water.
On deck, a man walks
diagonally toward the bow such that his path
forms an angle
θ
= 19
◦
with a line perpen
dicular to the boat’s direction of motion. He
walks at
v
m
= 2 m
/
s relative to the boat.
Draw the vectors to scale on a graph to
determine
the
answer.
θ
v
s
v
m
At what speed does he walk relative to the
water?
Correct answer: 6
.
91474 m
/
s (tolerance
±
5
%).
Explanation:
v
s
v
v
m
Scale: 1 m
φ
θ
α
β
Let :
v
s
= 6 m
/
s
v
m
= 2 m
/
s
v
= 6
.
91474 m
/
s
θ
= 71
◦
α
= 55
.
1286
◦
β
= 161
◦
,
and
φ
= 15
.
8714
◦
.
When you complete the parallelogram, the
resultant velocity
v
with respect to the water
is the side of the triangle opposite the obtuse
angle, which has a measure of
β
= 90
◦
+
θ
.
Let
vectorv
s
be the velocity of the ship,
vectorv
m
be the velocity of the man, and
vectorv
be the
resultant velocity of the man relative to the
water (Earth). By the law of cosines
v
2
=
v
2
m
+
v
2
s
−
2
v
m
v
s
cos
β
v
=
bracketleftbig
v
2
m
+
v
2
s
−
2
v
m
v
s
cos
β
bracketrightbig
1
/
2
=
bracketleftbig
(2 m
/
s)
2
+ (6 m
/
s)
2
−
2(2 m
/
s)(6 m
/
s) cos(109
◦
)]
1
/
2
=
bracketleftbig
(2 m
/
s)
2
+ (6 m
/
s)
2
+(7
.
81363 m
2
/
s
2
)
bracketrightbig
1
/
2
= 6
.
91474 m
/
s
.
Alternatively:
We can analyze the vector
addition using the components of the vectors.
Note:
vectorv
=
vectorv
s
+
vectorv
m
,
or
v
x
=
v
s
+
v
m
sin
θ
= (6 m
/
s) + (2 m
/
s) sin(19
◦
)
= 6
.
65114 m
/
s
and
v
y
=
v
m
cos
θ
= (2 m
/
s) cos(19
◦
)
= 1
.
89104 m
/
s
Hence the speed of the man with respect to
the water is
v
=
radicalBig
v
2
x
+
v
2
y
=
radicalBig
(6
.
65114 m
/
s)
2
+ (1
.
89104 m
/
s)
2
=
6
.
91474 m
/
s
.
Question 2, chap 3, sect 99.
part 2 of 2
10 points
At what angle to his intended path does the
man walk with respect to the water?
Correct answer: 55
.
1286
◦
(tolerance
±
5 %).
Explanation:
The law of sines can be used to compute the
requested angle
α
, which is the angle opposite
the ship’s path and velocity.
sin
α
v
s
=
sin
β
v
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homework 03 – ASH, BEN – Due: Feb 6 2008, 11:00 pm
2
sin
α
=
v
s
v
sin
β
α
= arcsin
bracketleftBig
v
s
v
sin
β
bracketrightBig
= arcsin
bracketleftbigg
6 m
/
s
6
.
91474 m
/
s
sin(109
◦
)
bracketrightbigg
= 55
.
1286
◦
.
Alternate Solution:
Using vector compo
nents from Part 1, we have
tan
φ
=
v
y
v
x
φ
= arctan
parenleftbigg
v
y
v
x
parenrightbigg
= arctan
parenleftbigg
1
.
89104 m
/
s
6
.
65114 m
/
s
parenrightbigg
= 15
.
8714
◦
.
Therefore the angle between
vectorv
m
and
vectorv
is
α
= 90
◦
−
θ
−
φ
= 90
◦
−
(19
◦
)
−
(15
.
8714
◦
)
=
55
.
1286
◦
.
Question 3, chap 3, sect 99.
part 1 of 2
10 points
A superhero flies 281 m from the top of
a tall building at an angle of 21
◦
below the
horizontal.
Draw the vectors to scale on a graph to
determine the answer.
a) What is the horizontal component of the
superhero’s displacement?
Correct answer:
262
.
336
m (tolerance
±
5
%).
Explanation:
262
.
336 m
−
100
.
701 m
281 m
Scale: 100 m
−
21
◦
Basic Concept:
∆
x
=
d
(cos
θ
)
Given:
d
= 281 m
θ
=
−
21
◦
Solution:
∆
x
= (281 m)[cos(
−
21
◦
)]
=
262
.
336 m
.
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 Spring '08
 ERSKINE/TSOI
 Acceleration, Work, Correct Answer, m/s

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