Teams A and B are participating in a 4 × 100 m relay race. At the start of the…

Concept: In a staggered-start timing problem, first pin down each runner's actual start and finish time using time = distance ÷ speed. To make two runners arrive together, set the required runner's finishing time equal to the target runner's finishing time — the speed needed is then simply that runner's leg distance divided by however much time is genuinely left before that shared target instant.

1. In a 4 × 100 m relay, each leg (each athlete's stretch) is 100 m long.

2. Team A's first athlete runs the first leg at 5 m/s, so he hands off to the second athlete at time = 100 ÷ 5 = 20 s from the start of the race.

3. Team A's second athlete also runs at 5 m/s, so he covers his 100 m leg in 100 ÷ 5 = 20 s, reaching the third athlete at 20 + 20 = 40 s. This 40 s mark is the time both teams' second athletes must reach together.

4. Team B's first athlete drops the baton and only restarts once team A's first athlete has covered 10 m; at 5 m/s that takes 10 ÷ 5 = 2 s, so team B's first athlete actually begins running 2 s after the race starts.

5. He also runs the first leg at 5 m/s, taking 100 ÷ 5 = 20 s, so he hands off to team B's second athlete at 2 + 20 = 22 s from the start of the race.

6. Team B's second athlete therefore has only 40 − 22 = 18 s left to cover his own 100 m leg and still reach the third athlete at the 40 s mark.

7. Required speed = distance ÷ time = 100 ÷ 18 = 50/9 m/s; converting to km/h by multiplying by 18/5 gives (50/9) × (18/5) = 20 km/h.

Check: starting his leg at 22 s and running at 20 km/h — the same as 100/18 m/s — team B's second athlete covers the 100 m leg in 100 ÷ (100/18) = 18 s, finishing at 22 + 18 = 40 s, exactly matching team A's second athlete's finishing time.

So the second athlete of team B must run at 20 km/h.