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

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