Question

1) A rural mail carrier leaves the post office and drives 20.0 km in a northerly direction. She then drives in a direction 60 degrees south of east for 40.0 km. Find her displacement from the post office using: a) the Tail-to tip method (use a neat graph with an appropriate scale) b) the Parallelogram method (use a neat graph with an appropriate scale) c) the Components method.

North D,760 0 Post office D,

Answer 1

OA^vector = O i^^+ 20 j^^ Vertical-bar AB^vector Vertical-bar = 40 OA^vector + AB^vector = OB^vector A rightarrow (0, 20) OB^vector = OA^vector + AB^vector B rightarrow (40 cos 60 degree, 20 - 40 sin 60) = (O i^^+ 20 j^^) +[(20 - 0) i^^+ (- 20(squareroot 2 - 1) - 20)j^^] B rightarrow (20, -20(-1 + squareroot 3)) OB^vector = 20 i^^- 20(squareroot 3 - 1)j^^ \r\n R^vector = OA^vector + OB^vector 40 = squareroot 20^2 + OB^2 + 2 times 20 times OB times cos theta OB = 24.78 OB^vector = 20 i^^- 20(squareroot 3 - 1)j^^OA^vector = 0 i^^+ 20 j^^OB^vector = > AB cos 60 degree i^^- (AB sin 60 - 20)j^^OB^vector = > 20 i^^- 20(squareroot 3 - 1) j^^ Vertical-bar OB^vector Vertical-bar = squareroot 20^2 + [20(squareroot 3 - 1)]^2 = 24.786

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