Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 47: 75

Answer

He should jump when the truck is approximately 1.5 poles away.

Work Step by Step

Agent Bond's initial position is considered to be $y=0$. We choose downward to be the positive direction. The initial velocity of Agent Bond is $v=0$ while his acceleration is $a=g$. His displacement if he jumped onto the truck would be: $y=15m-3.5m=11.5m$ Duration of Agent Bond's fall can be found by replacing $x$ with $y$ in the following formula: $$y=y_{0}+v_{0}t+\frac{1}{2}at^{2}$$ $$t=\sqrt (\frac{2y}{a})=\sqrt (\frac{2(11.5m)}{9.8m/s^{2}})=1.532s$$ Agent bond needs to jump when the truck is $d$ distance away. The truck is approaching at a constant velocity of $25m/s^2$. So we find $d$: $$d=vt=(25m/s^2)(1.532s)=38.3m$$ Converting this distance into ''poles'': $$(38.3m)(\frac{1pole}{25m})=1.532 poles\approx1.5 poles$$ So, Agent Bond should jump when the truck is approximately 1.5 poles away.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.