A train 75 m long overtook a person who was walking at the rate of 6 km/hr in the same direction and passed him in $7 \frac{1}{2}$ seconds. Subsequently, it overtook a second person and passed him in $6 \frac{3}{4}$ seconds. At what rate was the second person travelling? (M.A.T., 2008)

### Concept & Successive Relative Speeds Use the first overtaking scenario to find the train's absolute speed. Then, use that absolute speed in the second scenario to find the unknown speed of the second person. $$Speed_{relative} = \frac{Length_{train}}{Time}$$ ### Step-by-Step Solution 1. **Analyze First Person:** - Train length = 75 m, Time = $7.5$ s. - Relative speed 1 = $75 / 7.5 = 10$ m/s. - Convert to km/hr: $10 \times \frac{18}{5} = 36$ km/hr. - Since moving in the same direction, $Speed_{train} - 6 = 36 \implies Speed_{train} = 42$ km/hr. 2. **Analyze Second Person:** - Time = $6 \frac{3}{4}$ seconds = $27/4$ s = $6.75$ s. - Relative speed 2 = $\frac{75}{27/4} = \frac{300}{27} = \frac{100}{9}$ m/s. - Convert to km/hr: $\frac{100}{9} \times \frac{18}{5} = 40$ km/hr. - Since overtaking implies same direction, $Speed_{train} - Speed_{person2} = 40$. 3. **Solve for Second Person's Speed:** - $42 - Speed_{person2} = 40$ - $Speed_{person2} = 2$ km/hr. ### Exam Strategy & Shortcut Find both relative speeds in m/s first, convert to kmph, and track differences. Rel Speed 1: $10$ m/s = $36$ km/hr. Rel Speed 2: $100/9$ m/s = $40$ km/hr. Difference in relative speeds ($40 - 36 = 4$) must match the difference in the persons' speeds, but inversely. If relative speed increased by 4, the second person is moving 4 km/hr slower than the first. $6 - 4 = 2$ km/hr. ### Common Pitfall Miscalculating fractions when dividing by $6 \frac{3}{4}$. Always convert mixed fractions to improper fractions ($27/4$) before substituting into division operations. ### Final Answer Therefore, the correct answer is **2 km/hr**.