13. (15 points) One day, City $M$ was shrouded in heavy fog, and objects could only be seen clearly within 100 meters. Person A and Person B started from point $A$ on a straight road and walked in opposite directions at the same time. The speeds of A and B were 4 meters/second and 6 meters/second, respectively. After 1 minute, A reached point $B$, and B reached point $C$, then A and B turned around and walked back. After that, how long would it take for A and B to see each other? At that time, how far would A and B be from $A$ respectively?
【Analysis】1 minute $=60$ seconds, then after 1 minute, the two people are $(6+4) \times 60=600$ meters apart. Due to the fog, they can only see objects within 100 meters, so they will be able to see each other when they have walked $600-100=500$ meters back towards each other. At this point, they have walked $500 \div(6+4)=50$ seconds. Since person A walked for 60 seconds on the way out and has walked 50 seconds on the way back, with the speed unchanged, person A is now $4 \times(60-50)=40$ meters from point A. Similarly, person B is $6 \times(60-50)=60$ meters from point A.

【Solution】Solution: 1 minute $=60$ seconds,
$$
\begin{array}{l}
{[(6+4) \times 60-100] \div(6+4)} \\
=[10 \times 60-100] \div 10, \\
=[600-100] \div 10, \\
=50(\text { seconds); } \\
4 \times(60-50) \\
=4 \times 10, \\
=40 \text { (meters); } \\
6 \times(60-50) \\
=6 \times 10, \\
=60 \text { (meters). }
\end{array}
$$

Answer: After 50 seconds, person A and person B can see each other. At this time, person A is 40 meters from point A, and person B is 60 meters from point A.