Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Problems: 10

Answer

a) 1.35 km, 21.0$^{\circ}$ north of west b) 0.540 km/h

Work Step by Step

AE = BD – 0.50 = 2.15cos(35$^{\circ}$) - 0.50 $\approx$ 1.26 CE = CD – 0.75 = 2.15sin(35$^{\circ}$) - 0.75 $\approx$ 0.483 Displacement = AC = $\sqrt ((AE)^{2}+(CE)^{2})$ = $\sqrt (1.26^{2}+0.483^{2})$ $\approx$ 1.35 km $\theta$ = $tan^{-1}(\frac{CE}{AE})$ = $tan^{-1}(\frac{0.483}{1.26})$ $\approx$ 21.0$^{\circ}$ north of west Average velocity = $\frac{displacement}{time}$ = $\frac{1.35}{2.50}$ $\approx$ 0.540 km/h
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