Q.In a 1000 metres race, A, B and C get the gold, silver and bronze medals, respectively. If A beats B by 100 metres and B beats C by 100 metres, then by how many metres does A beat C?
Concept understanding — Relative Speed Ratio
Relative Speed Ratio — From Intuition to Precision
Imagine you're standing on a railway platform. A train passes you at 60 km/h. To you, it's moving fast. Now imagine you're sitting in another train running at 50 km/h in the same direction. The first train now seems to crawl past you at just 10 km/h. The same train, two different experiences — because your own motion changes how you perceive the other's speed.
That perceived speed is the relative speed. And when you compare two such relative speeds — say, the speed of one object relative to another, versus the speed of a third object relative to the same reference — you're dealing with a relative speed ratio.
Relative speed is always about one object as seen from another moving object. It is not the same as their individual speeds.
The Intuition First
Suppose two cars are on a straight road. Car A moves at 80 km/h, Car B at 60 km/h, both in the same direction. If you're in Car B, Car A approaches you at only 20 km/h. That's the relative speed: 80−60=20 km/h.
Now suppose Car C is coming from the opposite direction at 70 km/h. From Car B, Car C rushes toward you at 60+70=130 km/h.
The relative speed ratio compares two such relative speeds. For example, the ratio of Car A's speed relative to Car B, to Car C's speed relative to Car B, is 20:130=2:13.
That's all it is — a ratio of two relative speeds.
The Precise Statement
Let there be three objects (or two objects and a reference frame). Let vAB be the velocity of A relative to B, and vCB be the velocity of C relative to B. Then the relative speed ratio of A to C with respect to B is:
Relative Speed Ratio=∣vCB∣∣vAB∣
where ∣v∣ denotes speed (magnitude of velocity).
Relative Speed Ratio=speed of second object relative to same referencespeed of first object relative to reference
The reference object (B) is the one from whose perspective both speeds are measured.
Why It Matters
Relative speed ratios appear in:
- Time and distance problems — when two moving bodies meet or overtake, the ratio of their relative speeds determines the ratio of times or distances.
- River-boat problems — the ratio of the boat's speed relative to water to the river's speed relative to ground.
- Train-platform problems — comparing how fast two trains appear to cross a stationary observer.
In many exam problems, you don't need the absolute speeds — only the ratio of relative speeds. That ratio alone can give you the answer.
A Common Mistake
Students often confuse relative speed ratio with the ratio of the objects' own speeds. For example, if A moves at 80 km/h and C at 70 km/h, the ratio of their own speeds is 80:70=8:7. But if B is moving at 60 km/h in the same direction as A, the relative speed ratio is 20:130=2:13 — completely different.
Never substitute the objects' own speeds into a relative speed ratio. Always compute the relative speeds first.
One More Example
A train X is 200 m long and runs at 90 km/h. Train Y is 150 m long and runs at 72 km/h in the opposite direction. The relative speed of X with respect to Y is 90+72=162 km/h. The relative speed of Y with respect to X is also 162 km/h (same magnitude). So the relative speed ratio of X to Y (with respect to the ground) is 162:162=1:1 — they see each other at the same speed.
But if both ran in the same direction, the relative speed would be 90−72=18 km/h, and the ratio would be 18:18=1:1 again — because relative speed is symmetric in magnitude.
The ratio becomes interesting when you compare different pairs — like X relative to ground versus Y relative to ground, or X relative to Y versus Z relative to Y.
Final takeaway: Relative speed ratio is just a comparison of two apparent speeds from the same moving viewpoint. Compute each relative speed correctly, then take the ratio. That's the whole idea.
Since all three runners cover distances in the same ratio as their speeds over a given time, the distance C covers while A finishes the race can be found in two chained steps: first from A's pace to B's, then from B's pace to C's.
A beats C by 190 metres.
When A runs 1000 m, B runs 900 m; when B runs 900 m, C runs 810 m — so A beats C by 1000−810=190 m.
In a race, runners cover distances in the ratio of their speeds in the same time:
distance by Ydistance by X=speed of Yspeed of X (constant).
- A beats B by 100 m: when A covers 1000 m, B covers 1000−100=900 m. So AB=1000900.
- B beats C by 100 m: when B covers 1000 m, C covers 900 m. So BC=1000900.
- Distance C runs while B runs 900 m: C=900×1000900=1000810000=810 m.
- Thus when A runs the full 1000 m, B runs 900 m and C runs 810 m.
- A beats C by 1000−810=190 m.
A beats C by 190 metres.
- CBSE 2024Set 465/RQPS/41 markMCQQ.In a 1 km race, player P beats player Q by 18 metres or 9 seconds. What is P's time to complete the race ? (A) 512 seconds (B) 502 seconds (C) 491 seconds (D) 481 seconds
›Reveal solutionSolution
Q covers the gap of 18 m in 9 s (speed 2 m/s); P finishes the race exactly when Q reaches the 982 m mark, which takes 491 s.
speed=timedistance, and "P beats Q by d metres or t seconds" means Q runs the last d metres in t seconds.
- From "18 metres or 9 seconds", Q takes 9 s to run 18 m, so Q's speed is vQ=918=2 m/s.
- When P completes the full 1000 m, Q is 18 m behind, i.e. Q has run 1000−18=982 m.
- Both finish-of-P and Q-at-982 m happen at the same instant, so P's time equals the time Q needs for 982 m: t=2982=491 s.
✓Final answer(C) 491 seconds
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. By how much can A beat C in the same race ? (A) 32 m (B) 28 m (C) 24 m (D) 20 m
›Reveal solutionSolution
A:B = 100:75 and B:C = 100:96, so when A runs 100 m, C runs 72 m — A beats C by 28 m.
In a fixed-time race, distances are proportional: if A beats B by d1 in a race of length L, then distBdistA=L−d1L. Chain the ratios A:B and B:C to get A:C.
- A beats B by 25 m in 100 m: when A runs 100, B runs 100−25=75 m.
- B beats C by 4 m in 100 m: when B runs 100, C runs 100−4=96 m.
- Find C's distance while A runs 100 m (so B runs 75 m): C=75×10096=72 m.
- A beats C by 100−72=28 m.
✓Final answer(B) 28 m
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