Answer
At most $54.81$ $mi/h$
Work Step by Step
We are given the function $d(t)=v+\frac{v^2}{25}$
We have to find the range of speeds where the stopping distance $d(t)$ will not exceed $175ft.$, that is :
$v+\frac{v^2}{25}\leq175$, also note that the distance has to be non-negative value.
Simplify :
$v^2+25v-4375\leq0$
Let's solve the inequality using interval method
$v^2+25v-4375=0$
$D=b^2-4ac=625+17500=18125$
$v_1=\frac{-b-\sqrt{D}}{2a}=\frac{-25-25\sqrt{29}}{2}\approx -79.81$
$v_2=\frac{-b+\sqrt{D}}{2a}=\frac{-25+25\sqrt{29}}{2}\approx 54.81$
We have the following intervals (and the signs as follos) :
$(-\infty. -79.81]$ ; $[-79.81, 54.81]$ ; $[54.81, +\infty)$
Positive, Negative, Positive
As the speed cannot have a negative value, the range of speeds is in the interval $[0, 54.81]$
