An engine (train) of length 1000 m is moving at 10 m/s. A bird starts at the…

Concept: In a chase/relative-speed problem, two bodies moving toward each other (opposite directions) have relative speed equal to the SUM of their speeds, while a body chasing another moving the same way has relative speed equal to the DIFFERENCE of their speeds. Time to cover a fixed distance at a given relative speed = Distance ÷ Relative Speed.

Application: Let the bird's onward speed be x m/s and its return speed be 2x m/s; the engine itself keeps moving at 10 m/s throughout.

1. Onward leg (engine to far end): the bird flies against the engine's direction of travel while the far end advances to meet it — the two close from opposite directions, so their speeds ADD: relative speed = (x + 10) m/s, giving time = 1000/(x + 10) s.

2. Return leg (far end back to engine): the bird now flies in the same direction as the engine's travel, chasing the engine which keeps pulling away at 10 m/s — a same-direction pursuit, so the speeds SUBTRACT: relative speed = (2x − 10) m/s, giving time = 1000/(2x − 10) s.

3. Total time: 1000/(x + 10) + 1000/(2x − 10) = 187.5.

4. Clearing the denominators and simplifying: 3000x = 375x2 + 1875x − 18750, which reduces to x2 − 3x − 50 = 0.

5. Solving the quadratic: x = (3 + √209)/2 ≈ 8.728 m/s (to 3 decimal places), so 2x ≈ 17.456 m/s.

6. Converting to km/hr (× 3.6) using this standard 3-decimal rounding: x ≈ 31.4208 km/hr and 2x ≈ 62.8416 km/hr.

Cross-check: Substituting x = 8.728 m/s back: 1000/(8.728 + 10) + 1000/(2×8.728 − 10) = 1000/18.728 + 1000/7.456 ≈ 53.4 + 134.1 ≈ 187.5 s, matching the given total time and confirming x ≈ 31.4208 km/hr, 2x ≈ 62.8416 km/hr as the rounded pair.