Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 2 - Exercises and Problems - Page 30: 61

Answer

Please see the work below.

Work Step by Step

We can find the average speed as $v=\frac{140}{3.6}=38.89\frac{m}{s}$ (a) $v=\frac{v_f+ v_i}{2}$ We plug in the known values to obtain: $38.89=\frac{53+v_i}{2}$ $v_i=25\frac{m}{s}$ (b) We know that $a=\frac{\Delta V}{\Delta t}$ We plug in the known values to obtain: $a=\frac{53-25}{3.6}=7.84\frac{m}{s^2}$ Now, we can find the required distance traveled $d=\frac{v_f^2-v_i^2}{2a}$ $d=\frac{(53)^2-(0)^2}{2(7.84)}=180m$
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