Answer
$43~mi/h$
Work Step by Step
$speed_{1}=v$ and $speed_{2}=v+7.9~mi/h$
$time_{1}=t$ and $time_{2}=t-13~s$
$distance_{1}=distance_{2}=1.0~mi$
$time=\frac{distance}{speed}$
$time_{1}=\frac{distance_{1}}{speed_{1}}$
$t=\frac{1.0~mi}{v}$
$time_{2}=\frac{distance_{2}}{speed_{2}}$
$t-13~s=\frac{1.0~mi}{v+7.9~mi/h}$, but $13~s=\frac{13}{3600}~h$
$t=\frac{1.0~mi}{v+7.9~mi/h}+\frac{13}{3600}~h$
Now, we eliminate t:
$\frac{1.0~mi}{v}=\frac{1.0~mi}{v+7.9~mi/h}+\frac{13}{3600}~h$
$(3600~mi)(v+7.9~mi/h)=(3600~mi)v+(13~h)(v+7.9~mi/h)v$
$(13~h)(v^{2})+(102.7~mi)v-28440~mi^{2}/h=0$
$v=\frac{-102.7~mi+\sqrt {(102.7~mi)^{2}+1478880~mi^{2}}}{26~h}=43~mi/h$
$v=\frac{-102.7~mi-\sqrt {(102.7~mi)^{2}+1478880~mi^{2}}}{26~h}\lt0$. But, the speed must be greater than zero.
