Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 26

Answer

(a) Time, $t\approx 11.41$ $s$ (b) Acceleration, $a\approx 44.7$ $m/s^{2}$

Work Step by Step

(a) Given, $v=60.0$ $mi/h$ $=\frac{60\times1609.34}{3600}$ $m/s$ $\approx26.82$ $m/s$ $a=+2.35$ $m/s^{2}$ $v=u+at$ $26.82=0+2.35t$ $t=\frac{26.82}{2.35}$ $s$ $t\approx 11.41$ $s$ (b) Given, $v=60.0$ $mi/h$ $=\frac{60\times1609.34}{3600}$ $m/s$ $\approx26.82$ $m/s$ $t=0.600$ $s$ $v=u+at$ $26.82=0+0.6a$ $a=\frac{26.82}{0.6}$ $m/s^{2}$ $a\approx 44.7$ $m/s^{2}$
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