Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 2 - Kinematics in One Dimension - Problems: 30

Answer

Running at a constant velocity would take 2.81 s longer

Work Step by Step

$x=ut+\frac{1}{2}at^{2}$ When a = 0.0105 $(1609)=(16.58)t+\frac{1}{2}(0.0105)t^{2}$ $0.00525t^{2}+16.58t-1609=0$ Using the quadratic formula: $t=\frac{-(16.58)±\sqrt ((16.58)^{2}-4(0.00525)(-1609)}{2(0.00525)}$ $t\approx 94.23, t\approx -3252.33$ $t\approx 94.23\;s$ since $t\gt0$ When a = 0: $(1609)=(16.58)t+\frac{1}{2}(0)t^{2}$ $t=\frac{1609}{16.58}\approx 97.04\;s$ 97.04 - 94.23 = 2.81 s Hence, running at a constant velocity would take 2.81 s longer
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