Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - Pull it All Together - Page 702: Task 2

Answer

$t=\frac{7s-33}{s^2-6s}$

Work Step by Step

Using $d=rt$, with a distance of $5$ miles and a speed of $s\text{ mi/h}$, the time, $t_1$, for the travel on the flat stretch is $$\begin{aligned} d&=rt \\ 5&=st_1 \\ t_1&=\frac{5}{s} .\end{aligned}$$ With a distance of $1$ mile and the speed is doubled, then the time, $t_2$, for the travel down the hill is $$\begin{aligned} d&=rt \\ 1&=(2s)t_2 \\ t_2&=\frac{1}{2s} .\end{aligned}$$ With a distance of $3$ miles and the downhill speed is reduced by $12\text{ mi/h}$, then the time, $t_3$, for the travel in the last part is $$\begin{aligned} d&=rt \\ 3&=(2s-12)t_3 \\ t_3&=\frac{3}{2s-12} .\end{aligned}$$ Hence, the time, $t$, it takes for the rider to finish the race is $$\begin{aligned} t&=t_1+t_2+t_3 \\&=\frac{5}{s}+\frac{1}{2s}+\frac{3}{2s-12} \\&= \frac{7s-33}{s^2-6s} .\end{aligned}$$Hence, the time it takes for the rider to finish the race is $t=\frac{7s-33}{s^2-6s}$. Using $f(s)=t$, then the function that gives the time to finish the race is $$ t=\frac{7s-33}{s^2-6s} .$$
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