Time, Speed & Distance

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high importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Relative Speed

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⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Relative speed is how fast one moving thing closes the gap with another:

opposite directions: $S_{rel} = S_1 + S_2$ ; same direction: $S_{rel} = S_1 - S_2$ .

Use it whenever the question says cross, overtake, meet or catch up. The distance to cover is the combined lengths (two trains) or the current gap (chases). Two movers from the same spot drift apart at the sum or the difference of their speeds.

Add or subtract

Same direction → subtract the speeds (the gap closes slowly). Opposite directions → add them (the gap closes fast).

\text{same way: } S_1 - S_2 \qquad \text{opposite ways: } S_1 + S_2

A gap must shrink, so the relative speed is always the difference or sum of the sizes — never negative.

Rule: Read the directions first, then pick add or subtract. This one habit fixes most errors.

Meeting head-on

Two objects start towards each other from a gap D. They meet after

t = \frac{D}{S_1 + S_2}

Cars 150 km apart at 40 and 60 km/h meet in $\dfrac{150}{100} = 1.5$ hours — 60 km from the first town. The meeting POINT comes from one side alone: distance from A = $S_A \times t$ .

Each side also takes a fixed SHARE of the gap: A covers $\dfrac{S_A}{S_A + S_B}$ of it. At 40 and 60, A covers $\dfrac{40}{100} = 40\%$ — exactly $\dfrac{2}{5}$ of 150 km = 60 km, the same answer with no time computed.

Overtaking

A faster object closes a gap G at $S_1 - S_2$ : time = $\dfrac{G}{S_1 - S_2}$ .

For two trains the gap to clear is BOTH lengths together — the back of the faster train must pass the front of the slower. Trains of 200 m and 150 m at 63 and 45 km/h: relative speed 18 km/h = 5 m/s, gap 350 m, time $\dfrac{350}{5} = 70$ seconds.

Watch: A moving man is a POINT — only the train's own length counts. A train is LONG — add both lengths.

Train and a walking man

The man adds no length, only speed. Same direction → subtract his speed; opposite → add.

A 120 m train at 54 km/h passes a man walking the same way at 6 km/h: relative speed 48 km/h = $\dfrac{40}{3}$ m/s → $120 \div \frac{40}{3} = 9$ seconds.

Late starters

Someone leaves later. The gap is simply what the early mover covered alone.

A leaves at 9 a.m. at 30 km/h; B chases from 10 a.m. at 40 km/h. At 10 a.m. the gap is 30 km, closed at 10 km/h → 3 more hours → 1 p.m., at $40 \times 3 = 120$ km from the start.

Tip: Time everything from the second starter's departure; add the delay back at the end if asked.

The shuttle trick

A bird flies back and forth between two approaching trains until they meet. Do NOT sum the zigzags — the bird flies for the whole meeting time:

\text{distance} = \text{bird's speed} \times \text{meeting time}

Trains 100 km apart at 20 and 30 km/h meet in 2 hours; a bird at 60 km/h covers $60 \times 2 = 120$ km. One multiplication, no series.

Growing gaps

Two movers start from the SAME point. Opposite ways: the gap after t hours is $(S_1 + S_2)t$ . Same way: it is $(S_1 - S_2)t$ — the slower one falls behind at the difference.

At 40 and 50 km/h from one point: after 3 hours they are 270 km apart (opposite ways) or 30 km apart (same way).

Question types you will see

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

Type 1very common3 practice Q

Two vehicles moving towards each other

How to spot it:

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

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

Method

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

Why it works:

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

Try this

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

Show solution

$\dfrac{150}{40 + 60} = 1.5$ hours.
Meeting point: $40 \times 1.5 = 60$ km from the first town.

Answer

1.5 hours (60 km from the first town)

Type 2very common2 practice Q

Same direction catch-up or overtake

How to spot it:

A chase or an overtaking move in the same direction; the time to catch up or cross is asked.

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

Method

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

Why it works:

Seen from the slower object, the faster one approaches at the difference of speeds.

Try this

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

Show solution

$S_{rel} = 63 - 45 = 18$ km/h $= 5$ m/s.
Distance = $200 + 150 = 350$ m.
Time = $\dfrac{350}{5} = 70$ s.

Answer

70 seconds

Type 3common2 practice Q

Train crossing a moving man

How to spot it:

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

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

Method

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

Why it works:

The man's size adds no length; only his speed changes how fast the train passes him.

Try this

A 120 m long train running at 54 km/h crosses a man walking in the same direction at 6 km/h. Find the time taken.

Show solution

$S_{rel} = 54 - 6 = 48$ km/h $= \frac{40}{3}$ m/s.
Time = $120 \div \frac{40}{3} = 9$ s.

Answer

9 seconds

Type 4common2 practice Q

Delayed start or shuttle until meeting

How to spot it:

The second person starts later (gap = the head start), or a bird shuttles between two approaching vehicles.

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

Method

Delayed start: compute the head start, then chase at the difference of speeds.
Shuttle: find the meeting time first (gap ÷ sum), then multiply by the shuttle's speed.
Never sum the zigzags.

Why it works:

The shuttling object simply flies for the whole duration, whatever its path.

Try this

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

Show solution

Head start = $60 \times 2 = 120$ km.
Closed at $90 - 60 = 30$ km/h → 4 h.
Distance = $90 \times 4 = 360$ km.

Answer

360 km

Type 5common

Distance between two movers after a time

How to spot it:

Two people or vehicles start from the same place; how far apart they are after t hours is asked.

d = (S_1 + S_2)\,t \ \text{(opposite)}, \qquad d = (S_1 - S_2)\,t \ \text{(same way)}

Method

Fix the directions: opposite ways add the speeds, same way subtracts.
Multiply the (sum or difference) by the time.
State the distance apart, including units.

Why it works:

The gap itself grows at the relative speed.

Try this

Two cyclists start from the same point. One rides at 40 km/h, the other at 50 km/h. How far apart are they after 3 hours if they ride in opposite directions? And in the same direction?

Show solution

Opposite: $(40 + 50) \times 3 = 270$ km.
Same way: $(50 - 40) \times 3 = 30$ km.

Answer

270 km opposite ways; 30 km the same way

Formula sheet

Relative speed

S_{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}

Gap between two movers

d = (S_1 \pm S_2) \times t

Shortcuts that save time

⚡ Subtract for the same direction

Overtaking uses the difference; for two trains the distance is both lengths together.

Example

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

Show solution

$72 - 54 = 18$ km/h $= 5$ m/s.
Time = $\dfrac{200}{5} = 40$ s.

Answer

40 seconds

⚡ Add for opposite directions

The closing speed is the sum even when one side is just a walking man.

Example

A 250 m train at 45 km/h crosses a man walking at 5 km/h the opposite way. Find the crossing time.

Show solution

$45 + 5 = 50$ km/h $= \frac{125}{9}$ m/s.
Time = $250 \times \frac{9}{125} = 18$ s.

Answer

18 seconds

⚡ 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. Find the total distance the bird covers.

Show solution

Meeting time = $\dfrac{100}{50} = 2$ h.
Bird covers $60 \times 2 = 120$ km.

Answer

120 km

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding speeds when both move the same way.

Same direction subtracts; opposite directions add.

Mistake 02

Using only one train's length when two trains cross each other.

Two trains → the distance is L₁ + L₂.

Mistake 03

Applying relative speed to a stationary pole.

Against a pole the relative speed is just the train's own speed.

Mistake 04

Summing the endless zigzag of a shuttle.

Shuttle distance = its speed × the meeting time.

Mistake 05

Timing a delayed start from the wrong moment.

Work from the second starter's departure; the gap is the head start.

Quick revision

Read this the night before the exam.

Same way: subtract speeds; opposite ways: add.
Meet time = gap ÷ (sum); meeting point from either side's speed × time.
Overtake = gap ÷ (difference); two trains → both lengths.
Late starter: gap = distance the early one already covered.
Shuttle distance = speed × meeting time.
Two from one point: apart at (sum)t or (difference)t.

Practice: 12 questions

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

Topic test · 12 questions

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

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