7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{c}$ and $v_{2}=4 \mathrm{~m} / \mathrm{c}$. 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 did not change direction arrived at the final point of their journey $t_{2}=10$ minutes earlier than the person who turned around. Determine how much time $t_{1}$ passed from the start of the journey until the meeting. (15 points)
Answer: 30 min

Solution. The distance from point $B$ to the meeting place $S=v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{1}\left(t_{1}-t_{2}\right)$ (5 points). As a result, we get: $t_{1}=\frac{v_{1} t_{2}}{v_{1}-v_{2}}=\frac{6 \cdot 10}{6-4}=30$ min (5 points).