Q. 8 From the time the front of a train enters a platform, it takes 25 seconds for the back of the train
to leave the platform, while travelling at a constant speed of 54 km/h. At the same speed, it takes
14 seconds to pass a man running at 9 km/h in the same direction as the train. What is the length
of the train and that of the platform in meters, respectively?
The problem involves calculating the train’s length (L_train) and platform’s length (L_platform) using relative speeds and time. Train speed is 54 km/h (15 m/s), man’s speed is 9 km/h (2.5 m/s), platform crossing takes 25s, and passing the man takes 14s. Option (D) 175m and 200m is correct after converting units and solving equations.
Key Concepts
Relative speed governs crossing times: train + platform lengths at train speed for platform; train length at relative speed (train – man) for the man. Convert speeds: 54×518=15 m/s, 9×518=2.5 m/s.
Step-by-Step Solution
Equation 1 (Platform): Time = Ltrain+Lplatform15=25 → Ltrain+Lplatform=375 meters.
Equation 2 (Man): Relative speed = 15 – 2.5 = 12.5 m/s; Time = Ltrain12.5=14 → Ltrain=175 meters.
Substitute: 175+Lplatform=375 → Lplatform=200 meters. Matches option (D).
Option Analysis
| Option | Train (m) | Platform (m) | Check Platform (s) | Check Man (s) | Valid? |
|---|---|---|---|---|---|
| (A) | 210 | 140 | 35015=23.3 | 21012.5=16.8 | No |
| (B) | 162.5 | 187.5 | 35015=23.3 | 162.512.5=13 | No |
| (C) | 245 | 130 | 37515=25 | 24512.5=19.6 | No |
| (D) | 175 | 200 | 37515=25 | 17512.5=14 | Yes |
Physics Behind It
Train covers train + platform (375m) in 25s at 15 m/s for platforms; only train length at 12.5 m/s relative to man. These match real-world train problems emphasizing unit conversion and relative motion.
