Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 36

Answer

(a) The particle passes through the hoop at t = 3.0 seconds. (b) $v_{4y} = -8.0~m/s$

Work Step by Step

(a) The area under the $v_x$ versus time graph is equal to the horizontal displacement. We can find the time when x = 20 m. Between t = 0 and t = 3.0 s, the area under the $v_x$ versus time graph is equal to 20 m. Therefore, the particle passes through the hoop at t = 3.0 seconds. (b) At t = 3.0 s, $y = 0,$ because the particle passes through the hoop at this time. Between t = 0 and t = 3.0 s, the area under the $v_y$ versus t graph must be equal to -9.0 m. Therefore: $(\frac{1}{2})(\frac{3}{4}v_{4y})(3.0~s) = -9.0~m$ $v_{4y} = -8.0~m/s$
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