7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{s}$ and $v_{2}=4 \mathrm{~m} / \mathrm{s}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who turned around, upon returning to their starting point, waited for the second person. They waited for $t_{2}=10 \mathrm{min}$. Determine how much time $t_{1}$ passed from the start of the journey until the meeting. (15 points)
Answer: 20 min

Solution. Let $S$ be the distance between points A and B. We get $S=v_{1} t_{1}+v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{2} t_{1}+v_{2} t_{1}+v_{2} t_{2}$ (5 points). As a result, we get $t_{1}=\frac{v_{2} t_{2}}{v_{1}-v_{2}}=\frac{4 \cdot 10}{6-4}=20 \text{min}$ (5 points).