$3 \cdot 25$ On a two-lane road, to overtake car $B$, car $A$ must pass and maintain a distance of 30 feet. At this moment, car $C$ is coming from the opposite direction, 210 feet away from $A$ (as shown in the figure). If $B$ and $C$ maintain their original speeds of $40 \mathrm{mph}$ and $50 \mathrm{mph}$ respectively, to safely overtake, the speed that $A$ must increase is
(A) $30 \mathrm{mph}$.
(B) $10 \mathrm{mph}$.
(D) $15 \mathrm{mph}$.
(E) $3 \mathrm{mph}$.
(C) $5 \mathrm{mph}$.
[Solution] Let the increased speed of car $A$ be $r$, and the distance it travels to safely overtake car $B$ be $d$. Then, when car $A$ safely overtakes car $B$,
$$
\frac{d}{50+r}=\frac{d-\frac{30}{5280}}{40},
$$

and when car $A$ safely passes car $C$,
$$
\frac{d}{50+r}=\frac{\frac{210}{5280}-d}{50} \text {. }
$$

Solving the above two equations gives
$$
r=5, \quad d=\frac{110}{5280} \text {. }
$$

Therefore, the answer is $(C)$.