Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 7 - Section 7.3 - Direct Variation - Exercises - Page 285: 51

Answer

$225\;mi$.

Work Step by Step

Given values are Distance $d_1=90.0\;mi$. Time $t_1=1\frac{1}{2}\;h$. Let the speed is $s=\frac{d}{t}$ Substitute all the values. $s=\frac{90.0\;mi}{1\frac{1}{2}\;h}$ Distance $d_2$. Time $t_2=3\frac{3}{4}\;h$. Speed is same. $s=\frac{d_1}{t_1}=\frac{d_2}{t_2}$ Substitute all the values. $\Rightarrow \frac{90.0\;mi}{1\frac{1}{2}\;h}=\frac{d_2}{3\frac{3}{4}\;h}$ Simplify. $\Rightarrow \frac{90.0\;mi}{\frac{3}{2}\;h}=\frac{d_2}{\frac{15}{4}\;h}$ $\Rightarrow 225\;mi=d_2$.
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