1. **Problem statement:** Train A leaves a station 45 minutes before Train B. Both trains travel in the same direction with speeds 36 km/h (Train A) and 48 km/h (Train B). We want to find when Train B will catch up to Train A. 2. **Formula and concept:** When two objects move in the same direction, the time for the faster one to catch the slower one is given by: $$\text{Time} = \frac{\text{Distance lead}}{\text{Relative speed}}$$ 3. **Calculate the distance lead of Train A:** Train A travels for 45 minutes = $\frac{45}{60} = 0.75$ hours before Train B starts. Distance lead = speed of Train A $\times$ time lead = $36 \times 0.75 = 27$ km. 4. **Calculate the relative speed:** Relative speed = speed of Train B $-$ speed of Train A = $48 - 36 = 12$ km/h. 5. **Calculate the time for Train B to catch Train A:** $$\text{Time} = \frac{27}{12} = 2.25 \text{ hours}$$ 6. **Interpretation:** Train B will catch Train A 2.25 hours after Train B starts. 7. **Optional: Calculate the distance from the station where they meet:** Distance = speed of Train B $\times$ time after Train B starts = $48 \times 2.25 = 108$ km. **Final answer:** Train B catches Train A 2.25 hours after Train B starts, 108 km from the station.