A and B run a 1 km race. If A gives B a start of 50 meters, A wins by 14…

Concept

In a race between two competitors of constant speed, a start can be given either as a head-start distance or a head-start time, and the resulting winning margin (in distance or time) always translates into a linear relationship between the two racers' times to cover the full distance. Two independent race conditions between the same pair give two such equations, which can be solved together for both unknown times.

Application

1. Let a = time (in seconds) A takes to run 1000 m, and b = time B takes to run 1000 m.

2. Case 1 -- distance start: A gives B a 50 m head start, so B only needs to cover 1000 - 50 = 950 m. Time for B to cover 950 m = (950/1000) x b = 0.95b. Since A wins by 14 s, B's time exceeds A's time by 14: 0.95b - a = 14 ... (1)

3. Case 2 -- time start: A gives B a 22 s head start (B begins running 22 s before A). When B finishes the full 1000 m at time b (from B's own start), A has been running for (b - 22) s. Since B wins by 20 m, A has covered only 1000 - 20 = 980 m in that time: (1000/a)(b - 22) = 980 ... (2)

4. From (1): a = 0.95b - 14.

5. Substitute into (2): 1000(b - 22) = 980(0.95b - 14) => 1000b - 22000 = 931b - 13720 => 69b = 8280 => b = 120.

6. Substitute b = 120 back into a = 0.95b - 14: a = 0.95(120) - 14 = 114 - 14 = 100.

Cross-check

With a = 100 s and b = 120 s, A's speed is 1000/100 = 10 m/s and B's speed is 1000/120 = 8.33 m/s. Case 1: B needs 950/8.33 = 114 s to cover 950 m, which is exactly 14 s more than A's 100 s -- matches. Case 2: with a 22 s head start, A runs for 120 - 22 = 98 s and covers 98 x 10 = 980 m, exactly 20 m short of 1000 m -- matches.

Result

The time taken by A to run 1 km is 100 seconds.