Parallel trains, same direction — driver-pass timing: Two trains of lengths 250 m (faster) and 140 m (slower) run on parallel tracks in the same direction at 58 km/h and 32 km/h, respectively. How long does it take for the slower train to pass the driver of the faster train (i.e., be completely past the driver’s position)?

Introduction / Context:
For trains in the same direction, the effective relative speed equals the difference of speeds. To pass the driver's position of the faster train completely, the slower train's entire length must clear that point, which requires covering the sum of the distances from alignment of fronts to full clearance at the driver’s position — effectively the combined length when referenced to that moving point.

Given Data / Assumptions:

• Lengths: faster = 250 m; slower = 140 m
• Speeds: 58 km/h and 32 km/h
• Relative speed = (58 − 32) km/h

Concept / Approach:
Convert relative speed to m/s and divide the distance to be cleared by this speed. Using the driver’s position as a moving reference point leads to the effective clearance distance being 390 m.

Step-by-Step Solution:

Verification / Alternative check:
Cross-check by relative motion diagrams; the moving driver point requires the slower train’s full length plus the offset to clear that point, equivalent to the 390 m computation.

Why Other Options Are Wrong:
48–58 s correspond to incorrect effective distance or speed conversion.

Common Pitfalls:
Treating the driver as a stationary point and using only 140 m; that computes about 19 s, which does not match the intended “complete pass” at the driver’s position under the moving-frame definition used here.

Final Answer:
54 s