An engine of a train of length 1000 m is moving along a straight track at 10 m/s. A bird starts from the engine and flies along the train to the rear end at a speed of x m/s relative to the ground, then immediately flies back from the rear end to the engine at a speed of 2x m/s. If the total time taken for the bird to go from the engine to the rear end and back again is 187.5 s, what are the values of x and 2x in km/h?

Introduction / Context:
This is a relatively advanced relative motion problem involving a bird flying back and forth along a moving train. The length of the train, speed of the train, and total time for the two legs of the bird's journey are given. The question checks deeper understanding of how relative speeds work when the reference points themselves are in motion.

Given Data / Assumptions:

• Length of the train = 1000 m.
• Speed of the train = 10 m/s, moving forward along the track.
• Bird flies from engine to rear end at speed x m/s relative to the ground.
• Bird then flies back from rear end to engine at speed 2x m/s relative to the ground.
• Total time for both legs of the bird's journey = 187.5 s.
• All motions are along the straight line of the train.

Concept / Approach:
The key idea is to compute the relative speed between the bird and the part of the train it is moving towards on each leg. When the bird flies from the front towards the rear, it effectively moves backward while the rear is moving forward. When it comes back from rear to front, both bird and engine move forward, but at different speeds. Distances between the bird and the target ends are 1000 m for each leg in the ground frame. By setting up time expressions for each leg using these relative speeds and summing them, we can solve for x and then express x and 2x in km/h.

Step-by-Step Solution:
Let train speed = 10 m/s. First leg (engine to rear): bird moves backward relative to ground at speed x, rear moves forward at 10 m/s. Relative speed between bird and rear end = x + 10 m/s. Time for first leg t1 = 1000 / (x + 10). Second leg (rear to engine): bird moves forward at 2x m/s, engine moves forward at 10 m/s. Relative speed between bird and engine = 2x − 10 m/s (2x must be greater than 10). Time for second leg t2 = 1000 / (2x − 10). Total time t1 + t2 = 187.5 s gives 1000 / (x + 10) + 1000 / (2x − 10) = 187.5. Solving this equation yields x ≈ 8.7284 m/s. Then 2x ≈ 17.4568 m/s. Convert to km/h: x ≈ 8.7284 * 3.6 ≈ 31.4208 km/h and 2x ≈ 17.4568 * 3.6 ≈ 62.8416 km/h.

Verification / Alternative check:
Substitute x ≈ 8.7284 into t1 and t2. Then t1 = 1000 / (10 + 8.7284) and t2 = 1000 / (2 * 8.7284 − 10). Adding these numerically gives approximately 187.5 s, confirming that the values of x and 2x fit the total time constraint very closely. Hence the corresponding option with 31.4208 and 62.8416 km/h is correct.

Why Other Options Are Wrong:
If we pick 21.4 and 42.8, 25.2 and 50.4, or 33.12 and 66.24 km/h and convert them back to m/s, the resulting values of x will not satisfy the equation 1000 / (x + 10) + 1000 / (2x − 10) = 187.5. Their computed total times will differ significantly from 187.5 s, so they cannot be correct solutions.

Common Pitfalls:
Typical mistakes include assuming that the train is stationary, which ignores relative motion, or incorrectly setting relative speeds (for example, subtracting instead of adding in the first leg). Some also forget that distances between bird and train ends are 1000 m for both legs in the ground frame. Careful diagramming of each leg and consistent use of relative speed formulas are essential.

Final Answer:
Thus, the correct pair of speeds is 31.4208 and 62.8416 km/h.