AFCAT Numerical Ability · Speed, Time and Distance

Trains: Lengths, Platforms and Crossings

A train covers its own length to pass a point, and its own length plus the other length to pass a platform or another train, at the relative speed when the other thing moves.

All Speed, Time and Distance notes

Why this matters

AFCAT has set eight of the chapter's past-year questions here, from 2015 to 2018. Almost every one is a choice of which lengths to add and whether the speeds add or subtract, with km/h turned into m/s on the way.

Concept 1 of 5: Passing a pole or a standing man

A pole, a signal post or a man standing still has no length. The train has passed it when it has moved forward by one train length.

Definition

Time to pass a point $= \dfrac{\text{train length}}{\text{train speed}}$ , with the speed in m/s.
Two trains passing the same standing man: each covers only its own length, at its own speed.
A speed in km/h: multiply by $\tfrac{5}{18}$ first.

Passing a point

t = \dfrac{L}{v}

Worked example

A 180 m train runs at 81 km/h. How long does it take to pass a signal post?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2015 · AFCAT (II) 2015 — Numerical Ability · 04-221Easy

Example 1 · Speed, Time and Distance · Trains

A train covers a distance of 12 km in 10 min. If it takes 6 s to pass a telegraph post, the length of the train is

Metres with metres per second

A 180 m train at 81 km/h:

180 \div 81

mixes metres with kilometres per hour. Change 81 km/h to 22.5 m/s first.

Drill 6 more on passing a pole or a standing man

Concept 2 of 5: Crossing a platform or a bridge

To clear a platform the front of the train must run the platform's length and then the train's own length, until the last coach leaves the far end.

Definition

Distance to cross $=$ train length $+$ platform length.
$t = \dfrac{L + P}{v}$ , $v = \dfrac{L + P}{t}$ , and $P = vt - L$ .
'A platform twice as long as the train': the distance is $3L$ .
A man who walks across a bridge gives you the bridge's length: his speed times his time.

Crossing a platform

t = \dfrac{L + P}{v}

Worked example

A 200 m train at 90 km/h crosses a 300 m platform. How long does it take?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Speed, Time and Distance · Trains

A train

240 \text{ m}

long crosses a bridge

360 \text{ m}

long in

40 \text{ seconds}

. What is the speed of the train?

Add the train's own length

A 200 m train crossing a 300 m platform must cover 500 m. Using 300 m stops the clock while most of the train is still on the platform.

Drill 13 more on crossing a platform or a bridge

Concept 3 of 5: Length and speed from two crossings

Two crossings at the same speed give two equations. Subtract them and the train's length drops out: the extra time was spent on the extra length.

Definition

A pole in $t_1$ and a platform $P$ in $t_2$ : $v = \dfrac{P}{t_2 - t_1}$ , then $L = v t_1$ .
Two platforms $P_1, P_2$ in $t_1, t_2$ : $v = \dfrac{P_2 - P_1}{t_2 - t_1}$ .
Then $L = vt - P$ from either crossing.
A man standing on the platform counts as a pole.

Speed from two crossings

v = \dfrac{P_2 - P_1}{t_2 - t_1}

Worked example

A train passes a pole in 12 s and a 240 m platform in 28 s. Find its speed and length.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2015 · AFCAT (I) 2015 — Numerical Ability · 04-223Easy

Example 3 · Speed, Time and Distance · Trains

A train travelling at a uniform speed clears a platform 200 m long in 10 s and passes a telegraph post in 5 s. The speed of the train is

The man on the platform is a pole

Passing a man who stands on the platform takes the train's length only. Adding the platform to that crossing double counts it.

Drill 16 more on length and speed from two crossings

Concept 4 of 5: Passing someone who is moving

A walker or cyclist is still a point, so the train covers only its own length, but at the relative speed from the last page: the difference if they go the same way, the sum if they meet.

Definition

Same direction: $t = \dfrac{L}{v - u}$ . Opposite: $t = \dfrac{L}{v + u}$ .
A man sitting in another train moves with that train. Use the two trains' relative speed and only the length of the train he watches.
Same direction in $t_1$ and opposite in $t_2$ : $(v - u)t_1 = (v + u)t_2$ , both equal to $L$ .
One crossing time, with length and speed both unknown, cannot fix a second crossing time.

Passing a moving person

t = \dfrac{L}{v \mp u}

Worked example

A 240 m train at 63 km/h passes a man walking at 9 km/h. How long does it take if he walks (a) the same way, (b) towards the train?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2015 · AFCAT (II) 2015 — Numerical Ability · 04-220Easy

Example 4 · Speed, Time and Distance · Trains

A train 150 m long is running with a speed of 68 km/h. In what time will it pass a man who is running at 8 km/h in the same direction in which the train is going?

The man in the train

A man in one train watching the other go past sees only the other train's length go by. Adding both trains' lengths is for when the trains clear each other completely.

Drill 18 more on passing someone who is moving

Concept 5 of 5: Two trains crossing or overtaking

When two trains pass each other completely, the fronts must separate by both lengths, and that gap opens at the relative speed.

Definition

Opposite directions: $t = \dfrac{L_1 + L_2}{v_1 + v_2}$ . Same direction (overtaking): $t = \dfrac{L_1 + L_2}{v_1 - v_2}$ .
Equal lengths with pole times $t_1, t_2$ : the speeds are $\tfrac{L}{t_1}, \tfrac{L}{t_2}$ , and crossing head on takes $\dfrac{2 t_1 t_2}{t_1 + t_2}$ .
An unknown second speed: relative speed $= \dfrac{L_1 + L_2}{t}$ , then take away the known speed.

Two trains

t = \dfrac{L_1 + L_2}{v_1 \pm v_2}

Worked example

Trains 190 m and 230 m long run towards each other at 50 km/h and 58 km/h. How long do they take to cross?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2016 · AFCAT (I) 2016 — Numerical Ability · 04-217Moderate

Example 5 · Speed, Time and Distance · Trains

A 150 m long train crosses a milestone in 15 s and a train of the same length coming from the opposite direction in 12 s. The speed of the other train is

Both lengths, and subtract for the same direction

Clearing another train covers

L_1 + L_2

, not one length. And overtaking on a parallel track closes at the difference of the speeds, which is why it takes so much longer.

Drill 21 more on two trains crossing or overtaking

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Passing a pole or a standing man
Passing a point
$t = \dfrac{L}{v}$
Crossing a platform or a bridge
Crossing a platform
$t = \dfrac{L + P}{v}$
Length and speed from two crossings
Speed from two crossings
$v = \dfrac{P_2 - P_1}{t_2 - t_1}$
Passing someone who is moving
Passing a moving person
$t = \dfrac{L}{v \mp u}$
Two trains crossing or overtaking
Two trains
$t = \dfrac{L_1 + L_2}{v_1 \pm v_2}$

Watch out for (5)

Metres with metres per second
→ Passing a pole or a standing man
Add the train's own length
→ Crossing a platform or a bridge
The man on the platform is a pole
→ Length and speed from two crossings
The man in the train
→ Passing someone who is moving
Both lengths, and subtract for the same direction
→ Two trains crossing or overtaking

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