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Chapter 3
Page 1
CHAPTER 3  Kinematics in Two Dimensions; Vectors
1.
We choose the west and south coordinate system shown.
For the components of the resultant we have
R
W
=
D
1
+
D
2
cos 45°
= (200 km) + (80 km) cos 45°
=
257 km;
R
S
= 0 +
D
2
= 0 + (80 km) sin 45°
=
57 km.
We find the resultant displacement from
R
= (
R
W
2
+
R
S
2
)
1/2
= [(257 km)
2
+ (57 km)
2
]
1/2
=
263 km
;
tan
=
R
S
/
R
W
= (57 km)/(257 km) = 0.222, which gives
= 13° S of W
.
2.
We choose the north and east coordinate system shown.
For the components of the resultant we have
R
E
=
D
2
= 10 blocks;
R
N
=
D
1
–
D
3
= 18 blocks – 16 blocks =
2 blocks.
We find the resultant displacement from
R
= (
R
E
2
+
R
N
2
)
1/2
= [(10 blocks)
2
+ (2 blocks)
2
]
1/2
=
10 blocks
;
tan
=
R
N
/
R
E
= (2 blocks)/(10 blocks) = 0.20, which gives
= 11° N of E
.
3.
From Fig. 3–6c, if we write the equivalent vector addition, we have
V
1
+
V
wrong
=
V
2
,
or
V
wrong
=
V
2
–
V
1
.
4.
We find the vector from
V
= (
V
x
2
+
V
y
2
)
1/2
= [(8.80)
2
+ (– 6.40)
2
]
1/2
=
10.9
;
tan
=
V
y
/
V
x
= (– 6.40)/(8.80) = – 0.727, which gives
= 36.0°
below the
x
axis
.
5.
The resultant is
13.6 m, 18° N of E
.
D
1
S
W
D
2
x
R
D
2
y
R
x
R
y
D
2
D
1
N
E
R
D
2
D
3
V
y
V
V
x
y
x
V
R
V
1
V
2
V
3
North
East
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View Full DocumentChapter 3
Page 2
6.
(
a
)
V
1
x
= – 6.0,
V
1
y
= 0
;
V
2
x
=
V
2
cos 45° = 4.5 cos 45° = 3.18 =
3.2
,
V
2
y
=
V
2
sin 45° = 4.5 sin 45° = 3.18 =
3.2
.
(
b
) For the components of the sum we have
R
x
=
V
1
x
+
V
2
x
= – 6.0 + 3.18 = – 2.82;
R
y
=
V
1
y
+
V
2
y
= 0 + 3.18 = 3.18.
We find the resultant from
R
= (
R
x
2
+
R
y
2
)
1/2
= [(– 2.82)
2
+ (3.18)
2
]
1/2
=
4.3
;
tan
=
R
y
/
R
x
= (3.18)/(2.82) = 1.13, which gives
= 48° above –
x
axis
.
Note that we have used the magnitude of
R
x
for the angle indicated on the diagram.
7.
(
a
)
(
b
) For the components of the vector we have
V
x
= –
V
cos
= – 14.3 cos 34.8° =
– 11.7
;
V
y
=
V
sin
=
14.3 sin 34.8° =
8.16
.
(
c
)
We find the vector from
V
= (
V
x
2
+
V
y
2
)
1/2
= [(– 11.7)
2
+ (8.16)
2
]
1/2
=
14.3
;
tan
=
V
y
/
V
x
= (8.16)/(11.7) = 1.42, which gives
= 34.9° above –
x
axis
.
This is within significant figures.
Note that we have used the
magnitude of
V
x
for the angle
indicated on the diagram.
8.
Because the vectors are parallel, the direction can
be indicated by the sign.
(
a
)
C
=
A
+
B
= 6.8 + (– 5.5)
=
1.3 in the
+
x
direction
.
(
b
)
C
=
A
–
B
= 6.8 – (– 5.5)
=
12.3 in the
+
x
direction
.
(
c
)
C
=
B
–
A
= – 5.5 – (6.8)
= – 12.3 in the
+
x
direction
or
12.3 in the
–
x
direction
.
9.
(
a
) Using the given angle, we find the components from
V
N
=
V
cos 38.5° = (635 km/h) cos 41.5° =
476 km/h
;
V
W
=
V
sin 38.5° = (635 km/h) sin 41.5° =
421 km/h
.
(
b
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 Fall '08
 Mueller

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