$4 \cdot 207$ Car A is 100 meters west of Car B. Now both cars start moving east at the same time. If Car A moves forward at a constant speed of 50 meters/second, and Car B accelerates forward at 20 meters/second, after how many seconds will the distance between the two cars be the smallest? At that time, how many meters apart are the two cars?
[Solution] Let the distance between the two cars be $S$ meters after $t$ seconds.
At this time, car A is $S_{1}=v t=50 t$ meters from the starting point, and car B is
$$
S_{2}=100+\frac{1}{2} a t^{2}=100+10 t^{2} \text {, }
$$

away from car A's starting point.

Therefore,
$$
\begin{aligned}
S & =S_{2}-S_{1}=100+10 t^{2}-50 t \\
& =10\left(t-\frac{5}{2}\right)^{2}+\frac{75}{2} .
\end{aligned}
$$

So when $t=\frac{5}{2}$ (seconds), $S$ is minimized, at which point, the distance between the two cars is $\frac{75}{2}$ meters.