Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Check Your Understanding - Page 35: 12


$\sqrt 6$ $v$

Work Step by Step

$v^{2}=u^{2}+2ax$ sub $u=0$ $v^{2}=2ax$ $v=\sqrt (2ax)$ $v_{new}$ =$\sqrt (2a(3x))$ =$\sqrt (6ax)$ = $\sqrt 6\times\sqrt (2ax)$ =$\sqrt 6\times\ v$
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