Time, Speed & Distance

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Trains Crossing Poles, Platforms & Trains

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⏱ 5 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A train covers its OWN length to pass a point, and its length PLUS the obstacle's to pass anything long.

cross a pole or man: $t = \dfrac{L_{train}}{S}$ ; cross a platform or another train: $t = \dfrac{L_1 + L_2}{S_{rel}}$ .

A pole crossing hands you the train's length for free: $L = S \times t$ . Compare the pole time with the platform time and the platform length falls out.

What crossing means

A train "crosses" an object when its front reaches the object and its back leaves it. So the distance travelled during a crossing is:

pole, man, signal — a POINT → the train's own length $L$ ;
platform, bridge, tunnel — LONG → $L$ + platform length;
another train — LONG → $L_1 + L_2$ .

Then $t = \dfrac{\text{distance}}{\text{speed}}$ , using the relative speed when both move.

Rule: Point object → one length. Long object → two lengths. Decide this before touching the numbers.

A pole hands you the train's length

t = \frac{L}{S} \qquad \text{so} \qquad L = S \times t

A 180 m train crossing a pole in 9 s runs at $\dfrac{180}{9} = 20$ m/s = 72 km/h. Any question that gives a pole crossing is giving you the length-speed pair for free.

Platforms and bridges

t = \frac{L + P}{S}

Best route: get S (or L) from the pole crossing, then use the platform crossing for the rest. A train crosses a pole in 12 s and a 270 m platform in 30 s. In the extra $30 - 12 = 18$ s it covers exactly the extra 270 m → $S = 15$ m/s = 54 km/h, and $L = 15 \times 12 = 180$ m.

Tip: Subtract the two times — the difference covers exactly the platform. Two equations become one division.

Two trains

Opposite directions: $t = \dfrac{L_1 + L_2}{S_1 + S_2}$ — closes fast.
Same direction: $t = \dfrac{L_1 + L_2}{|S_1 - S_2|}$ — closes slowly.

Equal-length trains crossing in 12 s (opposite) and 36 s (same way) hide a neat ratio: the speeds are in the ratio $\dfrac{36 + 12}{36 - 12} = 2 : 1$ .

Picture the front and the back

Draw the moment the crossing STARTS (front touches the object) and the moment it ENDS (back leaves it). The travel between those two pictures is the crossing distance — every formula above is just that distance measured.

A 300 m train entering a 700 m tunnel: at the start the nose is at the entrance; at the end the tail is at the exit. The nose has travelled $700 + 300 = 1000$ m. At 20 m/s that is 50 seconds — the platform rule with a roof.

Tip: When a question feels twisted, sketch the two moments. The distance between them is always L, L + P, or L₁ + L₂.

Two crossings, two unknowns

Neither length nor speed given? Two crossings give two equations. Bridges of 200 m and 400 m crossed in 20 s and 30 s:

\dfrac{L + 200}{S} = 20, \qquad \frac{L + 400}{S} = 30

Subtract: $\dfrac{200}{S} = 10$ → $S = 20$ m/s, then $L = 20 \times 20 - 200 = 200$ m. The subtract-first step drops the train length out of the way.

A man sitting in a train

A man on board train A is still just a point. When train B passes him, only B's length counts, at the relative speed.

Trains at 42 and 48 km/h run opposite ways. A man in the first watches the 160 m second train go past: relative speed $90$ km/h $= 25$ m/s → $\dfrac{160}{25} = 6.4$ s.

Watch: "A man in the other train" swaps which length is used — the crossing train's, not his own.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Train crossing a pole or a man standing

How to spot it:

A single train crosses a pole, a standing man or a signal post; length, speed or time is asked.

L = S \times t

Method

A pole is a point: the train covers only its own length.
Any two of L, S, t give the third.
Convert with 5/18 when the units disagree.

Why it works:

Crossing a point starts when the engine reaches it and ends when the last carriage leaves it — exactly one train-length.

Try this

A 210 m long train crosses a pole in 12 seconds. Find the length of a platform it crosses in 30 seconds at the same speed.

Show solution

$S = \frac{210}{12} = 17.5$ m/s.
Total in 30 s: $17.5 \times 30 = 525$ m.
Platform = $525 - 210 = 315$ m.

Answer

315 m

Type 2very common2 practice Q

Train crossing a platform or bridge

How to spot it:

A train crosses a structure of given length; the time, speed or structure length is asked, often paired with a pole crossing.

t = \frac{L + P}{S}, \qquad S = \frac{P}{t_{platform} - t_{pole}}

Method

Write the distance as L + P — the train travels its own length plus the platform's.
With a pole crossing too, subtract the times: the extra time covers exactly the platform.
Then L = S × pole time.

Why it works:

During the extra time between the crossings, the front of the train travels exactly one platform-length more.

Try this

A train crosses a 180 m platform in 20 seconds and a pole in 8 seconds. Find the length of the train.

Show solution

Extra 12 s covers 180 m → $S = \frac{180}{12} = 15$ m/s.
$L = 15 \times 8 = 120$ m.

Answer

120 m

Type 3very common4 practice Q

Two trains crossing each other

How to spot it:

Two trains cross — opposite or same direction; the time or a speed ratio is asked.

t = \frac{L_1 + L_2}{S_1 + S_2} \ (\text{opposite}), \qquad t = \frac{L_1 + L_2}{|S_1 - S_2|} \ (\text{same})

Method

The distance is always the SUM of both lengths.
Opposite → add speeds; same → subtract.
Equal lengths crossing in $t_o$ and $t_s$ : speed ratio $= \frac{t_s + t_o}{t_s - t_o}$ .

Why it works:

The ratio shortcut follows from $L_1 + L_2$ being the same in both runs, so speeds invert as the times.

Try this

Two trains of equal length cross each other completely in 12 seconds in opposite directions and in 36 seconds in the same direction. Find the ratio of their speeds.

Show solution

Ratio = $\dfrac{36 + 12}{36 - 12} = \frac{48}{24}$ .
= $2 : 1$ .

Answer

2 : 1

Type 4common2 practice Q

Two crossings give two unknowns

How to spot it:

Neither the train's length nor its speed is given; two different crossings (two bridges, or a pole and a platform) supply two equations.

\frac{L + A}{S} = t_1, \qquad \frac{L + B}{S} = t_2

Method

Set up one equation per crossing.
Subtract them: $\dfrac{B - A}{t_2 - t_1} = S$ — the length drops out.
Back-substitute for L.

Why it works:

The two structures' length difference is covered in the two times' difference.

Try this

A train crosses bridges of lengths 200 m and 400 m in 20 seconds and 30 seconds respectively. Find the length of the train.

Show solution

$S = \frac{400 - 200}{30 - 20} = 20$ m/s.
$L = 20 \times 20 - 200 = 200$ m.

Answer

200 m

Type 5occasional

A man sitting in the other train

How to spot it:

A passenger in one train is crossed by (or watches) another train; the time is asked.

t = \frac{L_{\text{crossing train}}}{S_1 \pm S_2}

Method

The passenger is a point: only the OTHER train's length passes him.
Opposite ways → add the speeds; same way → subtract.
Divide that length by the relative speed (in m/s).

Why it works:

The passenger's own train adds no length to the crossing distance.

Try this

Two trains 140 m and 160 m long run in opposite directions at 42 km/h and 48 km/h. A man sitting in the first train is crossed by the second. Find how long it takes.

Show solution

Relative speed = $42 + 48 = 90$ km/h $= 25$ m/s.
Distance = the second train's length = $160$ m.
Time = $\dfrac{160}{25} = 6.4$ s.

Answer

6.4 seconds

Formula sheet

Pole / man

L_{train} = S \times t

Platform / bridge

L_{train} + L_{platform} = S \times t

Train vs train

L_1 + L_2 = S_{rel} \times t

Two-equation extraction

S = \frac{P}{t_{platform} - t_{pole}}

Shortcuts that save time

⚡ Pole first: it is the train's own length

The pole crossing IS the train's length in motion.

Example

A 300 m long train crosses a pole in 15 seconds. Find its speed in km/h.

Show solution

$S = \frac{300}{15} = 20$ m/s.
$20 \times \frac{18}{5} = 72$ km/h.

Answer

72 km/h

⚡ Subtract the pole equation

Platform time minus pole time covers exactly the platform.

Example

A train at 72 km/h crosses a pole in 10 seconds and a platform in 30 seconds. Find the platform's length.

Show solution

Speed = $72 \times \frac{5}{18} = 20$ m/s; $L = 20 \times 10 = 200$ m.
$L + P = 20 \times 30 = 600$ m.
$P = 400$ m.

Answer

400 m

⚡ Platform crossing from rest

No pole given: just add both lengths and divide by the speed.

Example

A train 180 m long crosses a 220 m long platform in 25 seconds. Find its speed in km/h.

Show solution

$S = \frac{180 + 220}{25} = 16$ m/s.
$16 \times \frac{18}{5} = 57.6$ km/h.

Answer

57.6 km/h

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using only the platform's length when a train crosses it.

The train travels its own length plus the platform's: (L + P)/S.

Mistake 02

Adding a platform when crossing a pole.

A pole is a point — only the train's length counts.

Mistake 03

Leaving the speed in km/h while lengths are in metres.

Convert with × 5/18 before dividing.

Mistake 04

Adding speeds of two trains moving the same way.

Same direction subtracts the speeds.

Mistake 05

Using both lengths when a man sits in the other train.

A man is a point: only the crossing train's length moves past him.

Quick revision

Read this the night before the exam.

Point object: t = L/S; platform: t = (L + P)/S; two trains: t = (L₁ + L₂)/(S₁ ± S₂).
Pole in t_p and platform P in t: S = P/(t − t_p), L = S × t_p.
Two bridges: subtract the equations — difference of lengths ÷ difference of times = S.
Man in the other train: only the crossing train's length, at relative speed.
km/h → m/s × 5/18 before dividing any length.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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