SSC CGL Maths · Time, Speed and Distance

Relative Speed: Chasing and Meeting

Two movers on one road close or open the gap between them at the difference of their speeds when they go the same way, and at the sum of their speeds when they come towards each other.

All Time, Speed and Distance notes

Why this matters

Forty-two questions in the bank, the largest page of the chapter, and twenty-nine of them EASY. Most are a policeman chasing a thief, asked in one of four forms: how long, how far, what gap is left, or what speed. One idea, the gap closed at the relative speed, answers all four.

Concept 1 of 4: Closing a gap at the relative speed

Ride along with the thief and he seems to stand still, while the policeman comes on at only the extra speed he has. A policeman at 10 km/h behind a thief at 8 km/h gains 2 km every hour, so a gap of 1 km is gone in half an hour.

Definition

Same direction: the gap changes at the relative speed $s_1 - s_2$ , the faster speed minus the slower.
Time to catch $= \frac{\text{gap}}{s_1 - s_2}$ .
Gap left after a time $t$ : starting gap $- (s_1 - s_2) \times t$ .
Unknown speed or gap: read the same line backwards. Gap ÷ catch time gives the relative speed; add the slower speed to get the faster.
Units: with a gap in metres and speeds in km/h, work in metres per minute (1 km/h is $\frac{1000}{60}$ m a minute) or change to m/s with $\frac{5}{18}$ .
Two people who start together and walk the same way move apart at the same relative speed.

Catching up

t = \frac{\text{gap}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead

Worked example

A constable is 450 m behind a thief. The constable runs at 15 km/h and the thief at 12 km/h. How many minutes does the constable take to catch him?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 23 Sep 2024, Shift 1 · Q52Easy

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

A thief is noticed by a policeman from a distance of 92 m. The thief starts running and the policeman chases him. The thief and the policeman run at the speed of 90 km/h and 104.4 km/h, respectively. In how much time will the policeman catch the thief?

Adding the speeds in a chase

When both run the same way, the gap closes at the difference of the speeds. The sum is only for movers coming towards each other.

Dividing metres by km/h

A 450 m gap closed at 3 km/h does not take 150 of anything. Change the gap to km (0.45) or the speed to metres a minute (50) before you divide.

Drill 12 more on closing a gap at the relative speed

Concept 2 of 4: How far each one runs before the catch

Once you know how long the chase lasts, each runner's distance is his own speed times that time. The chaser runs further than the one ahead by exactly the starting gap.

Definition

Distance of the one ahead $= s_2 \times \frac{\text{gap}}{s_1 - s_2}$ ; chaser's distance $= s_1 \times \frac{\text{gap}}{s_1 - s_2}$ .
Chaser's distance minus the other's distance equals the starting gap: a quick check on any answer.
Fast route by ratio: in the same time the distances are in the ratio of the speeds. With speeds 5 : 4 the gap is 1 part, so the one ahead runs 4 gaps and the chaser 5 gaps. This is safe only when both run for the whole chase.
"Caught at a distance of" means from the chaser's starting point, unless the question names another point. Read whose start it names.
Speeds given only as a ratio are enough for the distances. You need the actual speeds only for the time.

Distance run before the catch

d_{\text{ahead}} = \frac{s_2 \times \text{gap}}{s_1 - s_2} \qquad d_{\text{chaser}} = \frac{s_1 \times \text{gap}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead

Worked example

A thief is 250 m ahead of a policeman. The thief runs at 14 km/h and the policeman at 16 km/h. How far does the thief run before he is caught?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 3 · Q71Easy

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

A thief is spotted by a policeman from a distance of 195 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief is 22 km/h and that of the policeman is 27 km/h, then how far would the thief have run (in m) before he is overtaken?

Giving the chaser's distance when the other's is asked

The two distances differ by the starting gap, and both are usually among the options. Underline whose distance the question wants before you pick.

Measuring the catch point from the wrong start

When a question asks at what distance the thief is caught, the distance is counted from the chaser's starting point, which is the gap behind the thief's starting point.

Drill 15 more on how far each one runs before the catch

Concept 3 of 4: A chase where the chaser starts later

When the chaser sets off late, the gap is not given: the one ahead builds it during the delay. Work out that lead first, and the rest is an ordinary chase.

Definition

Lead built during a delay $=$ the speed of the one ahead $\times$ the delay.
The catch takes $\frac{\text{lead}}{s_1 - s_2}$ after the chaser starts. For a clock answer, add it to the chaser's start time.
Both are at the same place at the catch, so measured from the common start their distances are equal.
The one ahead has been moving for the delay plus the chase time.
Unknown chaser speed: find where the one ahead is at a known later moment and how far the chaser has got by then; the chaser's distance over his own time is his speed.

Chase after a late start

t = \frac{s_2 \times \text{delay}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead
$t$ time from the chaser's start to the catch

Worked example

A cyclist leaves a town at 9:15 a.m. at 12 km/h. A motorcyclist leaves the same town at 10:45 a.m. along the same road at 30 km/h. When does the motorcyclist catch the cyclist?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 14 Jul 2023, Shift 2 · Q74Easy

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

A thief steals a van at 3:00 a.m. and drives it at a speed of 57 km/h. The thief is discovered at 4:00 a.m. and the owner starts the chase with another van at a speed of 76 km/h. At what time will he catch the thief?

Using the delay as the chase time

The delay only builds the lead. The chase itself lasts the lead divided by the relative speed, counted from the moment the chaser starts.

Adding the chase time to the wrong start

For a clock answer, add the chase time to the time the chaser set off, not to the time the one ahead set off.

Drill 5 more on a chase where the chaser starts later

Concept 4 of 4: Moving towards each other

Two cars driving at each other both eat into the same gap, so the gap shrinks at the sum of their speeds. Two cars 300 km apart at 70 km/h and 80 km/h close 150 km every hour and meet in 2 hours.

Definition

Towards each other: time to meet $= \frac{\text{distance apart}}{s_1 + s_2}$ .
One starts later: move the early starter on by its speed × the delay, close what is left at the sum of the speeds, and add the result to the later start time.
Sum and difference: two movers a distance $d$ apart who meet in $t_1$ going towards each other and in $t_2$ going the same way have $s_1 + s_2 = \frac{d}{t_1}$ and $s_1 - s_2 = \frac{d}{t_2}$ . Half the sum of these and half the difference give the two speeds.
Out and back: when one reaches the far end and turns back, by the time they meet the two together have covered twice the distance between the ends.
Fast route: starting together, the meeting point splits the distance in the ratio of the speeds.

Meeting head on

t = \frac{d}{s_1 + s_2} \qquad s_1 + s_2 = \frac{d}{t_1} \qquad s_1 - s_2 = \frac{d}{t_2}

$d$ distance between the two at the start
$t_1$ time to meet moving towards each other
$t_2$ time for the faster to catch the slower moving the same way

Worked example

Two towns are 335 km apart. A bus leaves the first at 1:15 p.m. at 50 km/h towards the second. A car leaves the second at 2:45 p.m. at 80 km/h towards the first. When do they meet?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 14 Sep 2025, Shift 1 · Q69Moderate

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

A train leaves Station A at 6:00 AM and travels towards Station B at a constant speed of 60 km/h. Another train leaves Station B at 8:00 AM and travels towards Station A at a constant speed of 90 km/h. The distance between the two stations is 540 km. At what time will the two trains meet?

Subtracting the speeds for movers coming towards each other

Head on, both close the gap, so the speeds add. The difference is only for one catching up with another going the same way.

Forgetting the early starter's lead

When one sets off an hour earlier, the distance still to close when the second starts is the full distance minus what the first covered in that hour.

Out and back: closing only the one-way distance

When one mover reaches the far end and turns back to meet the other, together they have covered twice the distance between the ends, not once.

Drill 6 more on moving towards each other

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 (4)

Closing a gap at the relative speed
Catching up
$t = \frac{\text{gap}}{s_1 - s_2}$
How far each one runs before the catch
Distance run before the catch
$d_{\text{ahead}} = \frac{s_2 \times \text{gap}}{s_1 - s_2} \qquad d_{\text{chaser}} = \frac{s_1 \times \text{gap}}{s_1 - s_2}$
A chase where the chaser starts later
Chase after a late start
$t = \frac{s_2 \times \text{delay}}{s_1 - s_2}$
Moving towards each other
Meeting head on
$t = \frac{d}{s_1 + s_2} \qquad s_1 + s_2 = \frac{d}{t_1} \qquad s_1 - s_2 = \frac{d}{t_2}$

Watch out for (9)

Adding the speeds in a chase
→ Closing a gap at the relative speed
Dividing metres by km/h
→ Closing a gap at the relative speed
Giving the chaser's distance when the other's is asked
→ How far each one runs before the catch
Measuring the catch point from the wrong start
→ How far each one runs before the catch
Using the delay as the chase time
→ A chase where the chaser starts later
Adding the chase time to the wrong start
→ A chase where the chaser starts later
Subtracting the speeds for movers coming towards each other
→ Moving towards each other
Forgetting the early starter's lead
→ Moving towards each other
Out and back: closing only the one-way distance
→ Moving towards each other

Test yourself on Time, Speed and Distance

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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