## Physics (10th Edition)

Published by Wiley

# Chapter 2 - Kinematics in One Dimension - Problems: 41

14 s

#### Work Step by Step

To find velocity when leaving the crossing: $x=vt-\frac{1}{2}at^{2}$ $v=\frac{x+\frac{1}{2}at^{2}}{t}$ $v=\frac{20.0+\frac{1}{2}(1.6)(2.4)^{2}}{(2.4)}\approx10.2533$ Time required to reach 32 m/s: $v=u+at$ $t=\frac{v-u}{a}=\frac{(32)-(10.2533)}{(1.6)}\approx14$ s

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