Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 13b


$distance = 3.4\times10^{2}$ $m$

Work Step by Step

They are travelling with an initial velocity of $37m/s $ and with a retardation of $2m/s^{2}$. Let $d$ be distance covered while slowing. We know that, $v^{2}=v_{o}^{2}+2a d$ We substitute the values in the formula above and solve: $0^{2}=37^{2}+2(-2)d$ $\frac{-37^{2}}{-4} = d$ $d =342.25$ $m$ $\approx 3.4\times10^{2}$ $metres$
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