Answer
$7.54$ miles/hour
Work Step by Step
Converting $94$ minutes to hours results in
$$
94\text{ minutes}\times\frac{1\text{ hour}}{60\text{ minutes}}=\frac{94}{60}=\frac{47}{30}\text{ hours}
$$
Let $x$ be the average speed during the second half of the race. Then $x+2$ is the average speed during the first half of the race.
Let $t$ be the time during the second half of the race. Since the total time in the race is $\frac{47}{30}$ hours, then the time in the first half of the race is $\left(\frac{47}{30}-t\right)$ hours.
Using $d=rt$ or the relationship of distance $(d)$, rate $(r)$, and time $(t)$ of an object in uniform motion, then the equation representing the first half of the race is
$$\begin{aligned}
d&=rt
\\
5&=(x+2)\left(\frac{47}{30}-t\right)
.&\text{(Eqn }1)
\end{aligned}$$
The equation representing the second half of the race is
$$\begin{aligned}
d&=rt
\\
5&=xt
\\
t&=\frac{5}{x}
.&\text{(Eqn }2)
\end{aligned}$$
Substituting Eqn $2$ into Eqn $1$ results in
$$\begin{aligned}
5&=(x+2)\left(\frac{47}{30}-\frac{5}{x}\right)
\\
5&=(x+2)\left(\frac{47x-150}{30x}\right)
\\
150x&=(x+2)(47x-150)
\\
47x^2-206x-300&=0
.\end{aligned}$$
Using $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
$$\begin{aligned}
x&=\frac{-(-206)\pm\sqrt{-206)^2-4(47)(-300)}}{2(47)}
\\&=
\frac{103\pm\sqrt{24709}}{47}
\end{aligned}
\\
\begin{array}{l|r}
x\approx5.54 & x\approx-1.15
\end{array}
$$Since $x$ should be nonnegative, then $x\approx5.54$. Hence, the average speed during the first half of the race, $x+2$, is approximately $5.54+2=7.54$ miles per hour.
