AFCAT Numerical Ability · Speed, Time and Distance

Relative Speed: Meeting, Separating and Chasing

Two moving things close or open the gap between them at the sum of their speeds when they go opposite ways, and at the difference when they go the same way.

All Speed, Time and Distance notes

Why this matters

AFCAT set this in 2018 and 2019, both EASY. Most of the page is one decision, add or subtract, but the HARD questions hide it behind a late start, a head start, or speeds worked out from the times after meeting.

Concept 1 of 6: Adding or subtracting the speeds

Sit on one of the two movers. The other seems to move at the sum of the speeds if you are going opposite ways, and at the difference if you are going the same way.

Definition

Opposite directions (towards each other or apart): relative speed $= a + b$ .
Same direction: relative speed $= a - b$ .
Gap after $t$ hours $=$ relative speed $\times t$ . Time to reach a gap $=$ gap $\div$ relative speed.
'One is 8 km/h faster' or 'one is $\tfrac35$ of the other': write both speeds with one unknown, then use the sum or difference.

Relative speed

v_{\text{rel}} = a + b \;(\text{opposite}), \qquad v_{\text{rel}} = a - b \;(\text{same way})

Worked example

Two cyclists leave the same point at 15 km/h and 12 km/h. How far apart are they after 2 h if they ride (a) the same way, (b) opposite ways?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

AFCAT · 2018 · AFCAT (I) 2018 — Numerical Ability · 04-207Easy

Example 1 · Speed, Time and Distance · Relative Speed and Meeting

Two trains start from a certain place on two parallel tracks in the same direction. The speeds of the trains are 45 km/h and 40 km/h, respectively. The distance between the two trains after 45 min will be

Moving apart still adds

Cars driving away from each other open the gap at the sum of their speeds, just as cars driving towards each other close it at the sum. Only the same direction subtracts.

Drill 17 more on adding or subtracting the speeds

Concept 2 of 6: Meeting head on: when and where

Two movers coming towards each other close the gap at the sum of their speeds. In the time it takes, each covers a share of the gap in proportion to its own speed.

Definition

Time to meet $= \dfrac{D}{a + b}$ .
Meeting point: distance from A's end $= a \times \text{time} = \dfrac{a}{a + b} D$ .
'One had gone $k$ km more when they met': time $= \dfrac{k}{a - b}$ , then $D = (a + b) \times \text{time}$ .
If they carry on after meeting, the gap opens again at $a + b$ .
Given the meeting time and $D$ , the sum of the speeds is $\tfrac{D}{t}$ .

Time and place of meeting

t = \dfrac{D}{a + b}, \qquad d_A = \dfrac{a}{a + b} D

Worked example

Two towns are 360 km apart. Trains leave them at the same time towards each other at 50 km/h and 70 km/h. When and where do they meet?

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The same idea in a real exam question:

AFCAT · 2019 · AFCAT (I) 2019 — Numerical Ability · 04-204Easy

Example 2 · Speed, Time and Distance · Relative Speed and Meeting

The distance between two platforms

x

and

y

is 500 km. One train starts from

x

towards

y

at 20 km/h and another train moves towards

x

from

y

at 30 km/h. What is the distance from

x

of the point where the two trains cross each other?

The faster one covers more of the gap

At 50 km/h and 70 km/h the 360 km is shared as 150 and 210, in the ratio of the speeds, not 180 each.

Drill 30 more on meeting head on: when and where

Concept 3 of 6: Meeting when one starts later

If one mover starts later, let the first run alone until the second sets off. Then the gap that is left closes at the sum of the speeds.

Definition

Distance run alone $=$ first speed $\times$ head time. Gap left $= D -$ that.
Time after the second starts $= \dfrac{\text{gap left}}{a + b}$ . Add the head time if the answer is counted from the first start.
Timetable questions: each speed is $D$ divided by that train's journey time. $D$ is unknown but cancels.

A late second start

t' = \dfrac{D - a\,t_0}{a + b}

Worked example

Towns P and Q are 400 km apart. A car leaves P at 7 am at 40 km/h; another leaves Q at 8 am at 50 km/h towards P. When do they meet?

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The same idea in a practice question:

Practice questionModerate

Example 3 · Speed, Time and Distance · Relative Speed and Meeting

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?

Count from the right start

The 4 h after 8 am is from the second car's start. If the question counts from the first start, add the hour it ran alone.

Drill 3 more on meeting when one starts later

Concept 4 of 6: Chasing: catching up from behind

A chaser gains on the one ahead at the difference of their speeds. The head start is the gap that this gain has to close.

Definition

Catch-up time $= \dfrac{\text{head start}}{v_{\text{chaser}} - v_{\text{leader}}}$ .
A head start in time becomes a gap: leader's speed $\times$ that time.
'How far does the thief run before he is caught': thief's speed $\times$ catch-up time. The chaser runs that plus the head start.
Speeds in the ratio $6 : 5$ : the chaser gains 1 part while the leader runs 5, so the leader runs $5 \times$ the head start.
A head start in metres and speeds in km/h: change the gain into m/s (2 km/h is $\tfrac59$ m/s).

Catch-up time

t = \dfrac{\text{head start}}{v_{\text{chaser}} - v_{\text{leader}}}

Worked example

A lorry leaves a town at 40 km/h. A car follows from the same place 1 h 30 min later at 60 km/h. When and where does it catch the lorry?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Speed, Time and Distance · Relative Speed and Meeting

A thief steals a car and drives at

80 \text{ km/h}

. A policeman starts chasing

2 \text{ hours}

later at

100 \text{ km/h}

. When will the policeman catch the thief?

The two runners cover different distances

The thief runs 450 m and the policeman 600 m: they differ by the head start. Read which one the question asks about.

A head start in time is not a distance

'Starts 1 h 30 min later' must be turned into a gap,

40 \times 1.5 = 60

km, before dividing by the gain.

Drill 11 more on chasing: catching up from behind

Concept 5 of 6: A bird flying to and fro

A bird flying back and forth between two trains never stops until the trains meet. So its distance is its speed times the time the trains take to meet; the zigzags do not matter.

Definition

Time for the two to meet $= \dfrac{D}{a + b}$ .
Bird's distance $=$ bird's speed $\times$ that time.
Where the bird starts, and which way it turns, makes no difference.

The bird's distance

\text{bird's distance} = v_{\text{bird}} \times \dfrac{D}{a + b}

Worked example

Two trains 150 km apart run towards each other at 40 km/h and 35 km/h. A bird flies between them at 60 km/h until they meet. How far does it fly?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 5 · Speed, Time and Distance · Relative Speed and Meeting

Two friends start from their homes

15

km apart and walk towards each other at

3

km/h and

2

km/h respectively. A dog starts with the first friend, runs to the second friend, then back to the first, and so on until they meet. If the dog's speed is

10

km/h, what distance does the dog cover?

Do not follow the zigzags

Adding up each leg of the bird's flight is an endless sum. One multiplication, its speed times the meeting time, gives the total.

Drill 5 more on a bird flying to and fro

Concept 6 of 6: Speed ratios from meeting and passing times

After two movers meet, each still has to cover the stretch the other covered before the meeting. So the times they take after meeting reveal the ratio of their speeds.

Definition

Two movers start together from A and B. After meeting, the one from A takes $t_1$ to reach B and the one from B takes $t_2$ to reach A. Then $\dfrac{v_A}{v_B} = \sqrt{\dfrac{t_2}{t_1}}$ .
They met $\sqrt{t_1 t_2}$ after starting.
The same gap closed in $t_{\text{opp}}$ going opposite ways and $t_{\text{same}}$ the same way: $\dfrac{a + b}{a - b} = \dfrac{t_{\text{same}}}{t_{\text{opp}}}$ .
One starts earlier and they meet at a new time: write both distances and equate them to the same total.

Times after meeting

\dfrac{v_A}{v_B} = \sqrt{\dfrac{t_2}{t_1}}, \qquad t_{\text{meet}} = \sqrt{t_1 t_2}

Worked example

Two trains start at the same time from X and Y towards each other. After meeting, the one from X takes 16 h to reach Y and the one from Y takes 25 h to reach X. Find the ratio of their speeds and when they met.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 6 · Speed, Time and Distance · Relative Speed and Meeting

Two trains start simultaneously from stations

A

and

B

towards each other. The first train reaches

B

9

hours after meeting the second train. The second train reaches

A

4

hours after meeting the first train. In how many hours after the start do they meet?

The quicker finish belongs to the faster train

With 16 h and 25 h after meeting, the train taking 16 h is the faster one, so the ratio is

5 : 4

and not

4 : 5

Drill 6 more on speed ratios from meeting and passing times

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 (6)

Adding or subtracting the speeds
Relative speed
$v_{\text{rel}} = a + b \;(\text{opposite}), \qquad v_{\text{rel}} = a - b \;(\text{same way})$
Meeting head on: when and where
Time and place of meeting
$t = \dfrac{D}{a + b}, \qquad d_A = \dfrac{a}{a + b} D$
Meeting when one starts later
A late second start
$t' = \dfrac{D - a\,t_0}{a + b}$
Chasing: catching up from behind
Catch-up time
$t = \dfrac{\text{head start}}{v_{\text{chaser}} - v_{\text{leader}}}$
A bird flying to and fro
The bird's distance
$\text{bird's distance} = v_{\text{bird}} \times \dfrac{D}{a + b}$
Speed ratios from meeting and passing times
Times after meeting
$\dfrac{v_A}{v_B} = \sqrt{\dfrac{t_2}{t_1}}, \qquad t_{\text{meet}} = \sqrt{t_1 t_2}$

Watch out for (7)

Moving apart still adds
→ Adding or subtracting the speeds
The faster one covers more of the gap
→ Meeting head on: when and where
Count from the right start
→ Meeting when one starts later
The two runners cover different distances
→ Chasing: catching up from behind
A head start in time is not a distance
→ Chasing: catching up from behind
Do not follow the zigzags
→ A bird flying to and fro
The quicker finish belongs to the faster train
→ Speed ratios from meeting and passing times

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