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: 31


$1.0\; m/s^{2}$ left

Work Step by Step

Right is the positive direction: $v^{2}=u^{2}+2ax$ $a=\frac{v^{2}-u^{2}}{2x}=\frac{0^{2}-5.0^{2}}{2(12.5)}=-1.0\; m/s^{2}$ Hence, acceleration is $1.0\; m/s^{2}$ left
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