homework 04 – FIERRO, JEFFREY – Due: Jan 28 2008, 11:00 pm
1
Question 1, chap 3, sect 3.
part 1 of 2
10 points
A boy runs 17
.
1 blocks North, 15
.
3 blocks
Northeast, and 1
.
6 blocks West.
Determine the length of the displacement
vector that goes from the starting point to his
final position.
Correct answer: 29
.
4014
(tolerance
±
1 %).
Explanation:
Let ˆ
ı
= East and ˆ
= North. The displace
ment vector is
vector
R
= 17
.
1 ˆ
+ 15
.
3 cos 45
◦
ˆ
ı
+ 15
.
3 sin 45
◦
ˆ
−
1
.
6 ˆ
ı
=
R
e
ˆ
ı
+
R
n
ˆ
.
The length of
vector
R
is

vector
R

=
radicalBig
R
2
e
+
R
2
n
=
radicalBig
(15
.
3 cos 45
◦
−
1
.
6)
2
+ (17
.
1 + 15
.
3 sin 45
◦
)
2
= 29
.
4014
.
Question 2, chap 3, sect 3.
part 2 of 2
10 points
Determine the direction of the displacement
vector.
(Use counterclockwise as the posi
tive angular direction, between the limits of
−
180
◦
and +180
◦
from East)
Correct answer: 71
.
7269
◦
(tolerance
±
1 %).
Explanation:
The reference angle is
θ
= arctan
vextendsingle
vextendsingle
vextendsingle
vextendsingle
R
n
R
e
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= 71
.
7269
◦
.
Since the displacement vector lies in quad
rant 1, the angle is
0 +
◦
71
.
7269
◦
= 71
.
7269
◦
.
Question 3, chap 3, sect 1.
part 1 of 2
10 points
A football player runs directly down the
field for 28 m before turning to the right at
an angle of 26
◦
from his original direction
and running an additional 11 m before being
tackled.
a) What is the magnitude of the runner’s
total displacement?
Correct answer:
38
.
1924
m (tolerance
±
1
%).
Explanation:
28 m
11 m
d
−
26
◦
Note:
Figure is not drawn to scale.
Basic Concepts:
Δ
x
=
d
(cos
θ
)
Δ
y
=
d
(sin
θ
)
Δ
x
total
= Δ
x
1
+ Δ
x
2
Δ
y
total
= Δ
y
1
+ Δ
y
2
d
total
=
radicalBig
(Δ
x
total
)
2
+ (Δ
y
total
)
2
Given:
d
1
= 28 m
θ
1
= 0
◦
d
2
= 11 m
θ
2
=
−
26
◦
Solution:
Δ
x
1
=
d
1
(cos
θ
1
)
= (28 m)(cos 0
◦
)
= 28 m
Δ
y
1
=
d
1
(sin
θ
1
)
= (28 m)(sin 0
◦
)
= 0 m
Δ
x
2
=
d
2
(cos
θ
2
)
= (11 m)[cos(
−
26
◦
)]
= 9
.
88673 m
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homework 04 – FIERRO, JEFFREY – Due: Jan 28 2008, 11:00 pm
2
Δ
y
2
=
d
2
(sin
θ
2
)
= (11 m)[sin(
−
26
◦
)]
=
−
4
.
82208 m
Δ
x
total
= Δ
x
1
+ Δ
x
2
= 28 m + 9
.
88673 m
= 37
.
8867 m
Δ
y
total
= Δ
y
1
+ Δ
y
2
= 0 m + (
−
4
.
82208 m)
=
−
4
.
82208 m
d
total
=
radicalBig
(Δ
x
total
)
2
+ (Δ
y
total
)
2
=
radicalBig
(37
.
8867 m)
2
+ (
−
4
.
82208 m)
2
= 38
.
1924 m
Question 4, chap 3, sect 1.
part 2 of 2
10 points
b) At what angle to his original displace
ment is his total displacement (with counter
clockwise positive)?
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 Spring '08
 Turner
 Vector Space, Work, Correct Answer, JEFFREY

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