← All topics🚆 TrainsMaths
Speed, Distance & Time · Topic 5 of 5

🚆 Trains

2 exam question types, fully solved

Train problems are ordinary with one twist: a train has length, so “crossing” means the whole train has to clear the object. The first job is always to decide what counts as the distance.

Decide the distance, then divide

• Crossing a pole / tree / standing man (no width): the train covers only its own length, so .

• Crossing a platform / bridge / tunnel (has length): add them, .

• Two trains crossing: the distance is the SUM of both lengths, . Use relative speed — if opposite, if same direction.

• Fix the units first: if the speed is in km/h but lengths are in metres and time in seconds, convert with . Handy: km/h m/s.

🚆Train Crossing LabWatch the tail — it decides the distance
🚆 Train length: 150 m
🛤 Platform length: 250 m
⚡ Speed: 36 km/h
⏱ 0.0 s — front just reached itplatform 250 mtailtrain 150 mcovered 0 m of 400 m (150 + 250)
⏱ drag time
Distance
400 m
150 + 250
Speed
10 m/s
36 × 5⁄18
Crossing time
40 s
400 ÷ 10
The engine reaching the far end is NOT enough — the tail must clear it too. So the train travels 150 + 250 = 400 m, taking 400 ÷ 10 = 40 s.

A 150 m train at 10 m/s crosses a 250 m platform: distance m, so time s.

📝Practice Questions

Q1A train 150 m long crosses a pole in 15 seconds. What is its speed in km/h?

Q2Two trains 180 m and 220 m long move in opposite directions at 72 km/h and 108 km/h. How long do they take to cross each other?

Q3A 200 m long train moving at 36 km/h crosses a platform in 40 seconds. Find the length of the platform.

📚

Real exam questions — Trains

2 question types · 9 solved examples from real SSC papers

Train problems are just speed = distance ÷ time with one twist: a train has length, so the first job is always to decide what counts as the distance.

How to solve this type
A train is not a point — it has length, so "crossing" means the whole train must clear the object. Two cases:
1. Crossing a pole, tree, pillar or standing man (negligible width): the train only covers its OWN length, so distance=train length\text{distance}=\text{train length}.
2. Crossing a platform, bridge or tunnel (it has length): the train covers its own length PLUS the object, so distance=train length+platform length\text{distance}=\text{train length}+\text{platform length}.
Then use the single formula distance=speed×time\text{distance}=\text{speed}\times\text{time}. Units must match: if speed is in km/h but length is in metres and time in seconds, first convert km/h to m/s using ×518\times\dfrac{5}{18} (and m/s back to km/h using ×185\times\dfrac{18}{5}).

A train 150 m long crosses a pole in 15 seconds. Find its speed in km/h.

A30 km/hB36 km/hC45 km/hD40 km/h

A 200 m long train runs at 72 km/h. How long does it take to cross an electric pole?

A10 secondsB15 secondsC12 secondsD8 seconds

A 150 m long train crosses a 250 m long platform in 20 seconds. Find the speed of the train in km/h.

A80 km/hB72 km/hC60 km/hD54 km/h

A train 130 m long moving at 45 km/h crosses a platform in 30 seconds. Find the length of the platform.

A245 mB250 mC200 mD220 m

A 100 m long train crosses a 400 m long bridge at a speed of 25 m/s. Find the time taken.

A16 sB25 sC20 sD18 s