Answer
\[
50 \mathrm{s} ; 5000 \mathrm{ft}
\]
Work Step by Step
\[
\begin{array}{l}
s(t)=\frac{a}{2} t^{2}+v_{0} t+s_{0} \\
v(t)=a t+v_{0}
\end{array}
\]
Using Formula (10) and Formula (11) from the book:
\[
s(t)=2 t^{2}
\]
$a=4, v_{0}=0, s_{0}=0$
\[
s(t)=50 t+2500
\]
For the truck: $a=0, v_{0}=50, s_{0}=2500$
We are looking for a time when positions are equal.
\begin{array}{l}
2 t^{2}=2500+50 t \\
\Leftrightarrow -2500-50 t+ 2 t^{2}=0 \\
\Rightarrow t=50 \mathrm{s}
\end{array}
Distance at this time:
\[
s(t)=50^{2} \cdot 2=5000 \mathrm{ft}
\]
