Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Exercises and Problems - Page 92: 53


Please see the work below.

Work Step by Step

We know that the acceleration of the train is given as $a=\mu g$ We plug in the known values to obtain: $a=(0.58)(9.8)=5.684\frac{m}{s^2}$ The required distance is given as $d=\frac{v^2}{2a}$ $d=\frac{(\frac{140}{3.6})^2}{(2)(5.684)}=130m$ Hence, the train will stop in time.
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