Time, Speed & Distance

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Subtopic 2 of 5·← Speed, Distance, Time & Unit Conversion·Trains Crossing Poles, Platforms & Trains →

Relative Speed

How fast one moving object closes the gap with another:

$\text{opposite directions: } S_{\text{rel}} = S_1 + S_2 \qquad \text{same direction: } S_{\text{rel}} = S_1 - S_2$

Use relative speed whenever the question is 'cross', 'overtake', 'meet' or 'catch up'. The distance to cover is the combined lengths (trains) or the initial gap (chases, meetings).

Meeting point: two objects start towards each other → meet time $= \frac{\text{gap}}{S_1 + S_2}$ , regardless of individual lengths.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

Relative speed in one line

Whenever two moving things matter at once — meeting, crossing, overtaking, catching up — work with the gap and the relative speed: $\text{same direction: } S_{\text{rel}} = S_1 - S_2 \qquad \text{opposite directions: } S_{\text{rel}} = S_1 + S_2$ Same direction means the gap closes slowly (subtract); opposite directions make it close fast (add). The distance to cover is the current gap — for two trains crossing each other, the gap to "cross completely" is the sum of their lengths.

Meeting problems (opposite directions)

Two objects start towards each other from a gap D. They meet after $t = \frac{D}{S_1 + S_2}$ 240 km apart at 40 and 40 km/h → 3 hours. The meeting point is then fixed by one side alone: distance from A's start $= S_A \times t$ . Speeds 40 and 60 from 300 km apart: meet after 3 hours, 120 km from A's end.

Overtaking (same direction)

A faster object closes a gap G at $S_1 - S_2$ : $t = \frac{G}{S_1 - S_2}$ For trains the gap to clear is the two lengths together (the back of the faster must pass the front of the slower). For a cyclist or a moving man being crossed, the man is a point — only the train's own length counts.

Train vs a walking man

The man's own size is zero. Only his speed enters: same direction → subtract it from the train's speed; opposite → add. A 120 m train at 54 km/h passing a man walking the same way at 6 km/h: relative speed 48 km/h $= \frac{40}{3}$ m/s → $120 \div \frac{40}{3} = 9$ seconds.

Delayed starts

Someone leaves later — the gap is simply the distance the first runner covered alone. A starts at 9 a.m. at 30 km/h; B starts at 10 a.m. at 40 km/h in pursuit: gap at 10 a.m. = 30 km, closed at 10 km/h → 3 hours → 1 p.m., at $40 \times 3 = 120$ km from the start.

The shuttle (bird) trick

A bird flies back and forth between two approaching trains "until they meet". Do NOT sum the zigzags — the bird is simply flying the whole time, so $\text{distance} = \text{bird's speed} \times \text{meeting time}$ Trains 100 km apart at 20 and 30 km/h meet in 2 h; a bird at 60 km/h covers $60 \times 2 = 120$ km. One line, no series.

Sanity habits

Opposite directions: think "sum" before touching numbers.
A gap must shrink: if your relative speed came out negative, you subtracted in the wrong order (take the difference of magnitudes).
Convert km/h to m/s before dividing metres.

Common traps

Adding speeds when both move the same way.
Using one train's length only, when two trains cross each other.
In delayed-start problems, timing from the wrong start moment.
Summing an endless zigzag instead of speed × meeting time.

Quick revision

Same way: subtract; opposite ways: add.
Meet time = gap ÷ (sum of speeds); meeting point from either side.
Overtake = gap ÷ (difference of speeds); two trains → both lengths.
Late starter: gap = distance already covered.
Shuttle distance = speed × meeting time.

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: Two vehicles moving towards each othervery common3 practice Q

How to spot it:

Two buses/trains/people start towards each other from a known distance; the meeting time (or meeting point) is asked.

t = \frac{\text{gap}}{S_1 + S_2}, \qquad d_{\text{from A}} = S_1 \times t

Add the speeds — opposite directions always sum.
Meeting time = initial gap ÷ combined speed.
For the meeting point, multiply one vehicle's speed by the meeting time.

Why: the two vehicles together swallow the gap at the sum of their speeds.

Example: Two towns are 150 km apart. A car leaves one town for the other at 40 km/h and at the same time another car leaves the opposite town at 60 km/h. After how long do they meet?

$\frac{150}{40 + 60} = 1.5$ hours (meeting point: 60 km from the first town).

Type 2: Same direction — catch up / overtakevery common2 practice Q

How to spot it:

A chase or an overtaking move in the same direction; the time to catch up or the crossing time of two trains is asked.

t = \frac{\text{gap}}{S_{\text{fast}} - S_{\text{slow}}}

Subtract the speeds — same direction always differences.
Gap = the stated distance, or the sum of both train lengths.
Time = gap ÷ relative speed.

Why: viewed from the slower object, the faster one approaches at the difference of speeds.

Example: Two trains of lengths 200 m and 150 m run in the same direction at 63 km/h and 45 km/h. The time the faster train takes to completely cross the slower one is:

$S_{\text{rel}} = 18$ km/h $= 5$ m/s; total length 350 m → $\frac{350}{5} = 70$ seconds.

Type 3: Train crossing a moving mancommon2 practice Q

How to spot it:

A train crosses a man walking/running/cycling — alongside or opposite; the crossing time is asked.

t = \frac{L_{\text{train}}}{S_{\text{train}} \pm S_{\text{man}}}

The man is a point: only the train's own length has to pass.
Opposite direction → add the speeds; same direction → subtract.
Convert to m/s, divide the train's length by the relative speed.

Why: the man's size adds no length, only his speed changes how fast the train passes him.

Example: A 120 m long train running at 54 km/h crosses a man walking in the same direction at 6 km/h in:

$S_{\text{rel}} = 48$ km/h $= \frac{40}{3}$ m/s → $120 \times \frac{3}{40} = 9$ seconds.

Type 4: Delayed start / shuttle-till-meetingcommon2 practice Q

How to spot it:

The second person starts later (gap = what the first already covered), or a bird/object shuttles between two approaching vehicles until they meet.

\text{gap} = S_{\text{first}} \times t_{\text{alone}}, \qquad \text{shuttle distance} = S_{\text{bird}} \times t_{\text{meet}}

Delayed start: compute the head start, then chase it at the difference of speeds; add the times if asked from the very beginning.
Shuttle: find the meeting time of the two vehicles first (sum of speeds), then multiply by the shuttle's speed.
Never sum the zigzags.

Why: the shuttling object simply flies for the whole duration, whatever its path looks like.

Example: A train leaves a station at 60 km/h. Two hours later a second train leaves the same station along the same track at 90 km/h. How far from the station does the second train catch the first?

Head start $= 120$ km; gap closes at 30 km/h → 4 h after the start of the second train. Distance $= 90 \times 4 = 360$ km.

Formulas

Relative speed

S_{\text{rel}} = S_1 \pm S_2

Meeting time

t = \frac{\text{initial gap}}{S_1 + S_2}

Overtake time

t = \frac{\text{gap or combined length}}{S_1 - S_2}

Time after meeting (opposite starts)

\frac{\text{speed of other}}{\text{own speed}} = \frac{\text{time to B from meet}}{\text{time to A from meet}}

Shortcut tricks

⚡ Subtract for the same direction

Overtaking uses the difference; the length is both trains together.

Example: Two trains run in the same direction at 72 km/h and 54 km/h. The faster train, 200 m long, completely crosses the slower one in:

$S_{\text{rel}} = 20 - 15 = 5$ m/s ⇒ $t = \frac{200}{5} = 40$ s (treating the slower train as a point for the faster's length — add its length if given).

⚡ Add for opposite directions

Closing speed is the sum even if one object is a walking man.

Example: A 250 m train at 45 km/h crosses a man walking at 5 km/h in the opposite direction in:

$S_{\text{rel}} = 50$ km/h $= \frac{125}{9}$ m/s ⇒ $t = 250 \times \frac{9}{125} = 18$ s.

⚡ Distance flown till meeting

Find the meeting time first; any third object's distance = its speed × that time.

Example: Two trains 100 km apart approach each other at 20 km/h and 30 km/h. A bird flying at 60 km/h shuttles between them until they meet. The total distance the bird covers is:

Meet time $= \frac{100}{50} = 2$ h ⇒ bird covers $60 \times 2 = 120$ km (no need to sum the shuttles!).

Where students lose marks

Adding speeds when both move the same way (must subtract).
Using only one train's length when two trains cross each other.
Applying relative speed to a stationary pole (relative speed = train's own speed).
Summing an infinite shuttle zigzag instead of multiplying speed by the meeting time.

Practice sets — 12 questions

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

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.