Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 7 - Exercises and Problems: 63

Answer

Please see the work below.

Work Step by Step

We know that $Power \space available =0.27\times 370hp=99.9hp\times 746\frac{W}{hp}=74.5KW$ Now $v=\frac{60mi}{h}=\frac{60\times 1609\frac{m}{mi}}{1h\times 3600\frac{s}{h}}=26.8\frac{m}{s}$ Now $P=\frac{\Delta KE}{t}$ This can be rearranged as: $t=\frac{\Delta KE}{P}$ $t=\frac{\frac{1}{2}mv^2}{P}$ We plug in the known values to obtain: $t=\frac{\frac{1}{2}1200(26.8)^2}{74.5\times 10^3}$ $t=5.8s$
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