Answer
$t=0.084 \mathrm{\ h}$
Work Step by Step
The magnitude of this $\vec{r}$ is $r=\sqrt{(2.5-32 t)^{2}+(4.0-46 t)^{2}} .$ We minimize this by taking a derivative and requiring it to equal zero - which leaves us with an equation for $t$
$$
\frac{d r}{d t}=\frac{1}{2} \frac{6286 t-528}{\sqrt{(2.5-32 t)^{2}+(4.0-46 t)^{2}}}=0
$$
which yields $t=0.084 \mathrm{\ h}$
