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CDS Mathematics · Time, Speed and Distance

Relative Speed: Chasing and Meeting

Two movers going the same way close their gap at the difference of their speeds; going towards each other, at the sum.

All Time, Speed and Distance notesCDS Mathematics strategy

Why this matters

Thirteen PYQs. Every chase or meeting reduces to one division — gap ÷ relative speed — once you account for a head start. Answer the quantity asked: the time, the clock time, or the distance from a particular end.

Concept 1 of 2: Chasing: the same direction

If the chaser is uu and the one ahead is vv, the gap shrinks by u−vu - v every hour. A head start in time becomes a head start in distance: speed × head-start time.

Definition

  • Time to catch up =gapu−v= \dfrac{\text{gap}}{u - v}.
  • A head start of hh hours at speed vv is a gap of vhvh.
  • Distance run by the chaser until catching up =u×= u\times that time =u⋅gapu−v= \dfrac{u\cdot\text{gap}}{u - v}.
  • Several chasers meeting at one instant: set each distance equal at that time.

Catching up

t=gapu−vt = \dfrac{\text{gap}}{u - v}

Worked example

A cyclist leaves at 1212 km/hr; 11 hour later a car follows at 3636 km/hr. How far from the start does the car catch him?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q43Moderate

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

A thief steals a car parked in a house and goes away with a speed of 40 kmph. The theft was discovered after half an hour and immediately the owner sets off in another car with a speed of 60 kmph. When will the owner meet the thief ?

After the theft or after the start?

The owner catches up one hour after setting off, which is one and a half hours after the theft. Options pair the right distance with the wrong reference time.
Drill 6 more on chasing: the same direction

Concept 2 of 2: Meeting: opposite directions

Two trains heading towards each other close the distance at the sum of their speeds. If one leaves earlier, first move it forward by its head start, then share the rest.

Definition

  • Time to meet =remaining distanceu+v= \dfrac{\text{remaining distance}}{u + v}.
  • A later start: subtract the distance the first covered alone.
  • After meeting, if they take t1t_1 and t2t_2 hours to finish, their speeds satisfy uv=t2t1\dfrac{u}{v} = \sqrt{\dfrac{t_2}{t_1}}.
  • 'Meet mid-route': each covers half; the difference in their times equals the difference in starts.

After meeting

uv=t2t1\dfrac uv = \sqrt{\dfrac{t_2}{t_1}}

Worked example

Two places are 500500 km apart. A train leaves the first at 8 a.m. at 5050 km/hr; another leaves the second at 10 a.m. at 7575 km/hr towards it. When do they meet?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q16Moderate

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

There are two stations X and Y, 1320 km apart. A train starts from station X at 6 a.m. and moves at an average speed of 60 km/hr. At 2 p.m. another train starts from Y towards X and moves at an average speed of 80 km/hr. When do they meet ?

Distance from which end?

'How far from Q' asks for the distance the train starting at Q has covered, not the one from P. Compute both; they must add to the total.
Drill 5 more on meeting: opposite directions

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 (2)

Watch out for (2)

Test yourself on Time, Speed and Distance

20 past CDS questions from this chapter, timed at 24 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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