Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.12 - Modeling Variation - 1.12 Exercises - Page 128: 41

Answer

$5.33$mi/h

Work Step by Step

As per the given problem, we have: $F=kAs^2$ Here, $k$ is defined as constant of proportionality Plug the given values in the above expression, we get $220=k(40)(5)^2$ $\implies k=\dfrac{220}{(40)(25)}=0.22$ When $F=175 lb$ ; $A=28 ft^2$ and $k=0.22$ The, we have $175=(0.22) (28)(s^2)$ $s^2=\sqrt {28.41}=5.33$mi/h
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