University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 95: 3.34

Answer

b) v_{x}= -7.07m/s v_{y}= -42.1m/s c)v=42.7m/s \theta = 80.5^{\circ} South of West

Work Step by Step

b} v_{p/e}=v_{p/a} + v_{a/e} v_{p/e} = (0,-35) + (-10cos45 , -10 sin45) v_{p/e} = (-7.07, -42.1) m/s Therefore, v_{x}= -7.07m/s v_{y}= -42.1m/s c)v =\sqrt x((-7.07)^{2}+ (-42.1)^{2}) v= 42.7m/s \theta = 42.1/7.07 = 80.5 ^{\circ} Therefore, \theta = 80.5 ^{\circ} South of West
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