Answer
$\frac{180}{19}\text{ mi/h}$
Work Step by Step
Using $d=rt$ or the relationship of distance $(d)$, rate $(r)$, and time $(t)$ of an object in uniform motion, with $t=\frac{45}{60}=\frac{3}{4}\text{ hour}$ and $r=8\text{ mi/h}$, then
$$\begin{aligned}
d&=rt
\\
d&=\frac{3}{4}*(8)
\\
&=6
.\end{aligned}$$Hence, the distance of the runner's route is $6$ miles.
With $d=6$ and $t=\frac{38}{60}=\frac{19}{30}$ hour, then
$$\begin{aligned}
d&=rt
\\
6&=r\left(\frac{19}{30}\right)
\\
r&=6\left(\frac{30}{19}\right)
\\&=
\frac{180}{19}
.\end{aligned}$$Hence, the speed, $r$, of the runner at the end of the season is $\frac{180}{19}\text{ mi/h}$.
