2. The distance between points A and B is 90 km. At 9:00, a bus left point A for point B at a speed of 60 km/h. Starting from 9:00, every 15 minutes, buses leave point B towards point A at a speed of 80 km/h. The bus that left point A, after traveling 45 km, reduces its speed to 20 km/h due to a breakdown and continues at this speed. A passenger in this bus is considering whether to continue the slow journey to point B or return to point A in the first passing bus. How much time does the passenger have to make a decision?
2. A bus leaving point A will travel 45 km in $\frac{45}{60}$ hours. During this time, a bus leaving point B at 9:00 will travel $\frac{45}{60} \cdot 80=60$ km and will be closer to point A than the bus leaving point A. A bus leaving point B at 9:15 will travel $\left(\frac{45}{60}-\frac{15}{60}\right) \cdot 80=40$ km and will be 5 km away from the bus leaving point A. Therefore, the time it will take for them to meet is $\frac{5}{20+80}=\frac{1}{20}$ hours. The passenger has 3 minutes to make a decision.