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
θ
=
R
S
/
R
W
= (57 km)/(257 km) = 0.222, which gives
θ
= 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
θ
=
R
N
/
R
E
= (2 blocks)/(10 blocks) = 0.20, which gives
θ
= 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
θ
=
V
y
/
V
x
= (– 6.40)/(8.80) = – 0.727, which gives
θ
= 36.0°
below the
x
-axis
.
5.
The resultant is
13.6 m, 18° N of E
.
θ
D
1
S
W
D
2
x
R
D
2
y
R
x
R
y
D
2
θ
D
1
N
E
R
D
2
D
3
V
y
V
V
x
θ
y
x
V
R
θ
V
1
V
2
V
3
North
East

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