Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 4 - Exercises and Problems - Page 69: 69


a) 60 m/s b) .672 m

Work Step by Step

a) We know that the change in speed is equal to the integral of the force divided by the mass. Thus, we find: $v=\frac{1}{.165}\int_0^{.0224}(-25+(1.25\times10^5)t-(5.58\times10^6)t^2)dt$ $v=60 \ m/s$ b) We know that if velocity is the single integral of this function, it follows that distance traveled is the double integral of force. Thus, we take the integral of the function that results when we take the integral of the force to obtain: $d=\frac{1}{.165}\int_0^{.0224}(-25t+(.625\times10^5)t^2-(1.86\times10^6)t^3)dt$ $d=.672 \ m$
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