Time, Speed & Distance

high importance~2 Q in Tier 120 formulas⚡ 15 shortcuts5 subtopics

Overview Learn (5)Formulas (20)Examples (5)Practice (63)

Subtopic 3 of 5·← Relative Speed·Boats & Streams →

Trains Crossing Poles, Platforms & Trains

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

$\text{cross pole/man: } t = \frac{L_{\text{train}}}{S} \qquad \text{cross platform/train: } t = \frac{L_1 + L_2}{S_{\text{rel}}}$

Pole crossing gives the train's length directly: $L = S \times t$ . Comparing pole time with platform time exposes the platform length.

Convert the speed to m/s first, set up the length equation, solve. Two equations (pole + platform) solve for both length and speed.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 13 practice Q

Detailed notes

What "crossing" means

A train "crosses" an object when its front end reaches the object and its back end leaves it. So the distance travelled during a crossing is always measured from the train's own front to its own back over the object:

pole, man, signal post, tree — a point object → distance = train's own length $L$ ;
platform, bridge, tunnel — distance = $L + \text{platform length}$ ;
another train — distance = $L_1 + L_2$ .

Then simply $t = \dfrac{\text{distance}}{\text{speed}}$ (or the relative speed, if both move).

Pole / man / signal: the train's own length

$t = \frac{L}{S} \qquad \text{or} \qquad L = S \times t$ A 180 m train crossing a pole in 9 s runs at $\frac{180}{9} = 20$ m/s $= 72$ km/h. Every train question that gives a pole crossing is handing you the train's length-speed pair for free.

Platform / bridge: add the platform

$t = \frac{L + P}{S}$ Two-step strategy: get $S$ (or $L$ ) from the pole crossing first, then use the platform crossing for the remaining unknown. 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. This "subtract the times, the difference is the platform" trick avoids writing two equations.

Train vs train

Opposite directions: $t = \dfrac{L_1 + L_2}{S_1 + S_2}$ — the gap closes fast.
Same direction: $t = \dfrac{L_1 + L_2}{|S_1 - S_2|}$ — slow, uses the difference.
Crossing a point on the other train (or a man on it): only the crossing train's own length is used, at the relative speed.

Two crossings → solve two unknowns

When neither $L$ nor $S$ is given directly, two crossings give two equations. Bridges of 200 m and 400 m crossed in 20 s and 30 s: $\frac{L + 200}{S} = 20, \qquad \frac{L + 400}{S} = 30$ Subtract: $\frac{200}{S} = 10$ → $S = 20$ m/s, $L = 20 \times 20 - 200 = 200$ m. Any two fixed structures work the same way — subtract first, the train length drops out.

Common traps

Using only the train's length when a platform is mentioned (or only the platform — both are wrong).
Forgetting the other train's length when two trains cross each other.
Dividing metres by a km/h speed — convert with $\frac{5}{18}$ first.
In same-direction crossings, adding the speeds instead of subtracting.

Quick revision

Point object: $t = L/S$ . Platform: $t = (L+P)/S$ . Two trains: $t = (L_1+L_2)/(S_1 \pm S_2)$ .
Pole in $t_p$ s and platform $P$ in $t$ s → $S = \frac{P}{t - t_p}$ , $L = S\, t_p$ .
Two bridges: subtract the two equations — $\Delta L / \Delta t = S$ .
km/h → m/s × 5/18 before dividing lengths.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Train crossing a pole / man / signalvery common2 practice Q

How to spot it:

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

L = S \times t

A pole is a point: the train covers only its own length.
Any two of $L$ , $S$ , $t$ determine the third: $L = St$ , $S = L/t$ , $t = L/S$ .
Convert with $\frac{5}{18}$ if the units disagree.

Why: crossing a point object begins when the engine reaches it and ends when the guard's van leaves it — exactly one train-length of travel.

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

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

Type 2: Train crossing a platform / bridge / tunnelvery common2 practice Q

How to spot it:

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

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

Write the distance as $L + P$ — the train must travel its own length plus the platform.
If a pole crossing is also given, subtract the two times: the extra time covers exactly the platform length → $S = \frac{P}{\Delta t}$ .
Then $L = S \times t_{\text{pole}}$ if needed.

Why: during the extra time between the two crossings, the engine end travels exactly one platform-length more.

Example: A train crosses a platform of length 180 m in 20 seconds and a pole in 8 seconds. The length of the train is:

$\frac{L + 180}{L/8} = 20$ → $8L + 1440 = 20L$ → $L = 120$ m. (Speed $= 15$ m/s.)

Type 3: Two trains crossing each othervery common4 practice Q

How to spot it:

Two trains cross each other — opposite or same direction; the crossing time or a length/speed ratio is asked.

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

Distance is always the SUM of both lengths.
Opposite directions → add the speeds; same direction → subtract.
If equal lengths cross in $t_o$ s opposite and $t_s$ s in the same direction, the speed ratio is $\frac{t_s + t_o}{t_s - t_o}$ .

Why: the last point follows from $x+y = v_o t_o$ and $x+y = v_s t_s$ , so $v_o/v_s = t_s/t_o$ , and $\frac{v_o - v_s}{v_o + v_s} = \frac{t_s - t_o}{t_s + t_o}$ .

Example: Two trains of equal length cross each other completely in 12 seconds when moving in opposite directions and in 36 seconds when moving in the same direction. The ratio of their speeds is:

$\frac{v_{\text{fast}}}{v_{\text{slow}}} = \frac{36 + 12}{36 - 12} = \frac{48}{24} = 2:1$ .

Type 4: Two crossings → two unknownscommon2 practice Q

How to spot it:

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

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

Set up one equation per crossing.
Subtract them — the unknown length $L$ cancels or drops to $\Delta L = S\,\Delta t$ .
Back-substitute to get $L$ .

Why: subtracting eliminates one unknown immediately; the structures' length difference is covered in the time difference.

Example: A train crosses two bridges of lengths 200 m and 400 m in 20 seconds and 30 seconds respectively. The length of the train is:

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

Formulas

Pole / man

L_{\text{train}} = S \times t

Platform / bridge

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

Train vs train

L_1 + L_2 = S_{\text{rel}} \times t

Two-equation extraction

\frac{t_{\text{platform}}}{t_{\text{pole}}} = \frac{L + P}{L}

Shortcut tricks

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

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

Example: A 300 m long train crosses a pole in 15 seconds. Its speed in km/h is:

$S = \frac{300}{15} = 20$ m/s $= 20 \times \frac{18}{5} = 72$ 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. The platform's length is:

$L = 20 \times 10 = 200$ m; $L + P = 20 \times 30 = 600$ ⇒ $P = 400$ m.

⚡ Man in the other train

A man is a point — only the crossing train's length matters.

Example: A train 180 m long crosses a 220 m long platform in 25 seconds. Its speed is:

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

Where students lose marks

Forgetting to include the train's own length when crossing a platform or another train.
Adding the platform length when crossing a pole.
Leaving the speed in km/h while lengths are in metres.
Using the relative speed of a stationary object as the train's speed MINUS zero incorrectly in same-direction cases (it is simply the train's speed).

Practice sets — 13 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 13 questions

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