lenoir (wml297) – homework 07 – Turner – (58220)
1
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001
(part 1 of 2) 10.0 points
A
particle
undergoes
three
displacements.
The first has a magnitude of 15 m and makes
an angle of 28
◦
with the positive
x
axis. The
second has a magnitude of 6
.
5 m and makes
an angle of 143
◦
with the positive
x
axis. (see
the figure below).
After the third displacement the particle
returns to its initial position.
143
◦
28
◦
15 m
6
.
5 m
Find the magnitude of the third displace-
ment.
Correct answer: 13
.
5956 m.
Explanation:
Given :
bardbl
vector
A
bardbl
= 15 m
,
θ
a
= 28
◦
,
bardbl
vector
B
bardbl
= 6
.
5 m
,
and
θ
B
= 143
◦
.
θ
C
θ
A
θ
B
A
B
C
−
C
Since
vector
A
+
vector
B
+
vector
C
= 0
,
we have
vector
C
=
−
vector
A
−
vector
B
.
The components of the third displacement
vector
C
are
C
x
=
−
A
x
−
B
x
=
−
A
cos
θ
A
−
B
cos
θ
b
=
−
(15 m) cos 28
◦
−
(6
.
5 m) cos 143
◦
=
−
(13
.
2442 m)
−
(
−
5
.
19113 m)
=
−
8
.
05308 m
and
C
y
=
−
A
y
−
B
x
=
−
A
sin
θ
A
−
B
sin
θ
b
=
−
(15 m) sin 28
◦
−
(6
.
5 m) sin 143
◦
=
−
(7
.
04207 m)
−
(3
.
9118 m)
=
−
10
.
9539 m
.
The magnitude of
vector
C
is
bardbl
vector
C
bardbl
=
radicalBig
C
2
x
+
C
2
y
=
radicalBig
(
−
8
.
05308 m)
2
+ (
−
10
.
9539 m)
2
=
13
.
5956 m
.
002
(part 2 of 2) 10.0 points
Find the angle of the third displacement (mea-
sured from the positive
x
axis, with counter-
clockwise positive within the limits of
−
180
◦
to +180
◦
).
Correct answer:
−
126
.
323
◦
.
Explanation:
tan
θ
C
=
C
y
C
x
θ
C
= arctan
bracketleftbigg
C
y
C
x
bracketrightbigg
= arctan
bracketleftbigg
(
−
10
.
9539 m)
(
−
8
.
05308 m)
bracketrightbigg
=
−
126
.
323
◦
.
003
(part 1 of 2) 10.0 points
Consider two vectors
vector
A
and
vector
B
and their re-
sultant
vector
A
+
vector
B
. The magnitudes of the vectors
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lenoir (wml297) – homework 07 – Turner – (58220)
2
vector
A
and
vector
B
are, respectively, 14
.
6 and 8
.
3 and
they act at 75
◦
to each other.

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- Spring '10
- Turner
- Vector Space, Correct Answer, Euclidean vector, Vector Motors, Lenoir
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