Lesson 59 of 100 | Quantitative Aptitude / Speed, Distance and Time / चाल, दूरी और समय
Relative speed and meeting or overtaking
Learning outcome
Find meeting and catching times from a correctly measured initial gap, direction and relative speed.
Concepts and assumptions
Relative speed describes how quickly separation changes. Treat travellers as points moving along one straight route, maintaining their stated speeds and directions without stops or acceleration until the required event.
When travellers approach each other, both reduce the gap. Closing speed = u + v, so meeting time = initial gap ÷ (u + v). Opposite directions alone do not guarantee a meeting: moving away from each other increases the gap.
For a faster traveller following a slower one, each time unit gains only the speed difference:
Closing speed = faster speed - slower speed.
Catching time = initial gap ÷ closing speed.
This requires a positive gap and closing speed. Equal speeds leave the gap unchanged; a slower follower falls farther behind. Neither produces a catch under unchanged conditions.
Measure the gap when both positions are compared on the same clock. With a delayed departure, calculate the earlier traveller’s lead first. Catching time then starts at the later departure, unless the question asks otherwise.
Each person’s distance still equals their own speed × time. Relative speed describes the gap, not either person’s travel distance. Use compatible units.
Worked examples
Example 1 — Approach. Two travellers start simultaneously from points 231 km apart, moving towards each other at 42 km/h and 35 km/h. Closing speed = 42 + 35 = 77 km/h. Meeting time = 231 ÷ 77 = 3 h.
Example 2 — Catch a leader. A walker at 1.5 m/s is 720 m ahead of a runner at 4.5 m/s, both moving in the same direction. Closing speed = 4.5 - 1.5 = 3 m/s. Time = 720 ÷ 3 = 240 s = 4 min. Check: runner distance = 4.5 × 240 = 1080 m; walker distance = 1.5 × 240 = 360 m; 720 + 360 = 1080.
Example 3 — Delayed departure. A cyclist leaves P at 16 km/h; another follows from P 45 minutes later at 24 km/h. Lead = 16 × 45/60 = 12 km. Closing speed = 24 - 16 = 8 km/h. Time from the second departure = 12/8 = 1.5 h. Location = 24 × 1.5 = 36 km from P. Check: the first travels for 0.75 + 1.5 = 2.25 h, covering 16 × 2.25 = 36 km.
Common mistakes
Do not divide the gap by the follower’s speed while the leader moves. Include the lead and identify the clock’s starting instant.
Practice questions
- Travellers start simultaneously 192 km apart, approaching at 54 km/h and 42 km/h. Find the meeting time.
- A runner at 5 m/s is 150 m ahead of a pursuer at 8 m/s, moving in the same direction. Find the catching time.
- A truck leaves at 45 km/h; another follows from the same point 30 minutes later at 60 km/h on the same route. Find catching time from the second departure and location.
- Two runners move in the same direction at 6 m/s each, one 300 m behind. When does the rear runner catch up?
Worked answers
- Closing speed = 54 + 42 = 96 km/h. Time = 192 ÷ 96 = 2 h. Distances 108 km and 84 km confirm the initial separation.
- Closing speed = 8 - 5 = 3 m/s. Time = 150 ÷ 3 = 50 s. The pursuer covers 400 m; the leader covers 250 m plus the initial 150 m.
- Lead = 45 × 30/60 = 22.5 km. Relative speed = 15 km/h, giving 22.5 ÷ 15 = 1.5 h. Location = 60 × 1.5 = 90 km from the common starting point.
- Relative speed = 6 - 6 = 0 m/s. The 300 m gap stays unchanged, so there is no catch. Division by zero cannot provide a catching time.