Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 3 - Section 3.4 - Constant Acceleration - Example - Page 40: 3.3

Answer

$83.5\ m$

Work Step by Step

We use the equation $x=v_{x0}t+\frac{1}{2}a_xt^2$ to get that the change in x is 79 meters. We use the equation $y=v_{y0}t+\frac{1}{2}a_yt^2$ to find that the change in y is 27 meters. Thus, we find that the value of r, the displacement, is: $r = \sqrt{79^2+27^2}\approx 83.5\ m$
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