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 - Problems - Page 44: 33

Answer

The speed of the last car will be 21 m/s.

Work Step by Step

$v_0 = 0 ~m/s$ $x_0 = 0 ~m$ To find the acceleration, we can use this equation: $v^2 = v_0^2 + 2a(x-x_0)$ $a = \frac{v^2}{2x} = \frac{18^2}{2\times180} = 0.9 ~m/s^2$ We need to find the speed of the train when the front of the train has travelled a distance of 255 meters. $v^2 = v_0^2 + 2a(x-x_0) = 2\times0.9\times255 = 459$ $v = \sqrt{459} = 21 ~m/s$ Note that the answer is rounded off to two significant digits.
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