Train problems are just speed-distance-time problems with one twist: the train has length. Unlike a car (which we treat as a point), a train takes extra time to fully pass something because of its own size. Once we internalize this, every train problem becomes a simple D = S × T problem.
The one concept that makes it all click: the “distance” a train covers while passing something includes the train’s own length.
Why the Train’s Length Matters
Think of it like this: when a train passes a pole, the front of the train reaches the pole first, but the train hasn’t “fully passed” until the tail clears the pole too. So the total distance traveled = the entire length of the train.
For a platform, the front has to travel the full platform length PLUS the train’s own length to completely clear the platform.
Case 1: Crossing a Pole or Standing Person
The simplest case. Distance = train length. Speed = train’s speed.
Example 1: Crossing a pole
A train 200 meters long crosses a pole in 10 seconds. Find its speed.
Distance = 200 m. Time = 10 s.
Speed = 200/10 = 20 m/s = 20 × 18/5 = 72 km/h
Case 2: Crossing a Platform or Bridge
Distance = train length + platform length. Speed = train’s speed.
Example 2: Crossing a platform
A train 150 m long runs at 54 km/h. How long does it take to cross a platform 300 m long?
Total distance = 150 + 300 = 450 m.
Speed = 54 × 5/18 = 15 m/s.
Time = 450/15 = 30 seconds
Example 3: Finding train length
A train crosses a 200 m platform in 25 seconds and a 500 m platform in 40 seconds. Find the speed and length of the train.
Let train length = L meters and speed = S m/s.
Crossing 200 m platform: (L + 200)/S = 25 → L + 200 = 25S … (i) Crossing 500 m platform: (L + 500)/S = 40 → L + 500 = 40S … (ii)
Subtract (i) from (ii): 300 = 15S → S = 20 m/s = 72 km/h
From (i): L = 25(20) - 200 = 500 - 200 = 300 m
This is a classic pattern — two platform crossings give us two equations with two unknowns.
Case 3: Two Trains Crossing Each Other
This is where relative speed kicks in. The total distance both trains need to cover to fully pass each other is the sum of their lengths (L₁ + L₂).
Opposite directions: Relative speed = S₁ + S₂ (they’re approaching each other).
Same direction: Relative speed = S₁ - S₂ (the faster one overtakes the slower one).
Example 4: Opposite directions
Two trains of lengths 150 m and 200 m are moving toward each other at 40 km/h and 32 km/h. How long do they take to cross each other?
Total distance = 150 + 200 = 350 m.
Relative speed = 40 + 32 = 72 km/h = 72 × 5/18 = 20 m/s.
Time = 350/20 = 17.5 seconds
Example 5: Same direction
A train 250 m long traveling at 60 km/h overtakes a train 300 m long traveling at 42 km/h. How long does it take to completely pass it?
Total distance = 250 + 300 = 550 m.
Relative speed = 60 - 42 = 18 km/h = 18 × 5/18 = 5 m/s.
Time = 550/5 = 110 seconds
Notice how same-direction crossing takes way longer than opposite-direction crossing. That’s because the relative speed is much smaller.
Case 4: Man on a Train vs. Man on a Platform
This is a subtle but important distinction.
Man standing on the platform: He’s stationary, so relative speed = train’s speed. Distance = train length. (Same as crossing a pole.)
Man sitting on another train: He’s moving! So we use relative speed between the trains. Distance = length of the train that’s passing him (not his own train).
Example 6: Man on another train
Train A (200 m long) at 60 km/h passes a man sitting in Train B going in the opposite direction at 40 km/h. How long does it take?
Distance = Train A’s length only = 200 m (the man is a point, we just need Train A to pass him).
Relative speed = 60 + 40 = 100 km/h = 100 × 5/18 = 250/9 m/s.
Time = 200 / (250/9) = 200 × 9/250 = 1800/250 = 7.2 seconds
Shortcut: The Speed Ratio Method
When a train crosses a pole in time t₁ and a platform of length L in time t₂:
Train length = L × t₁ / (t₂ - t₁)
This saves us from setting up simultaneous equations.
Example: A train crosses a pole in 10 seconds and a 300 m platform in 25 seconds.
Train length = 300 × 10 / (25 - 10) = 3000/15 = 200 m
Common Traps
- Forgetting unit conversion — speeds in km/h, lengths in meters. Always convert to consistent units (m/s is usually best).
- Using only one train’s length when two trains cross each other — distance is ALWAYS L₁ + L₂.
- Man on a moving train — don’t forget he has speed too. Use relative speed.
- Same direction vs. opposite direction — getting the relative speed wrong (add vs. subtract).
Common Exam Variations
- Find speed given crossing time and length.
- Find train length from two different crossing scenarios.
- Two trains crossing each other (same or opposite direction).
- “A train passes a man in 10 seconds and a platform in 20 seconds” — find both length and speed.
- “At what speed should a train travel to cross a bridge in X seconds?”
- Time for trains to completely cross each other.
Practice Problems
Problem 1: A train 300 m long crosses a bridge 200 m long in 25 seconds. Find the speed of the train in km/h.
Problem 2: Two trains of lengths 200 m and 250 m are traveling in the same direction at 72 km/h and 36 km/h. How long does the faster train take to completely pass the slower one?
Problem 3: A train passes a standing man in 8 seconds and a 264 m long platform in 20 seconds. Find the length and speed of the train.
Answers
Problem 1: Distance = 300 + 200 = 500 m. Speed = 500/25 = 20 m/s = 20 × 18/5 = 72 km/h.
Problem 2: Total distance = 200 + 250 = 450 m. Relative speed = 72 - 36 = 36 km/h = 36 × 5/18 = 10 m/s. Time = 450/10 = 45 seconds.
Problem 3: Let train length = L, speed = S. L/S = 8 and (L + 264)/S = 20. From first equation: L = 8S. Substituting: (8S + 264)/S = 20 → 8S + 264 = 20S → 12S = 264 → S = 22 m/s. L = 8 × 22 = 176 m. Speed = 22 × 18/5 = 79.2 km/h.