College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 148: 18


The plane travels a distance of 4.22 km

Work Step by Step

Note that the sum of the vertical forces is zero. Let east be the positive direction. We can find the net horizontal force: $\sum F_x = 1.800~kN-1.400~kN = 400~N$ We can find the acceleration: $\sum F_x = m~a_x$ $a_x = \frac{\sum F_x}{m}$ $a_x = \frac{400~N}{1160~kg}$ $a_x = 0.345~m/s^2$ We can find the distance the plane travels in the next $60.0~s$: $\Delta x = v_{0x}~t+\frac{1}{2}a_xt^2$ $\Delta x = (60.0~m/s)(60.0~s)+\frac{1}{2}(0.345~m/s^2)(60.0~s)^2$ $\Delta x = 4220~m = 4.22~km$ The plane travels a distance of 4.22 km
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