Two trains and the same bridge – second crossing time A train at 90 km/h crosses a bridge in 36 s. Another train that is 100 m shorter crosses the same bridge at 45 km/h. How long does the second train take?

Introduction / Context:
Here, two trains of different lengths and speeds cross the same bridge. The first crossing provides the bridge length implicitly; then reuse that with the second train’s speed and length to find its time. Algebraic elimination avoids explicitly solving for individual lengths.

Given Data / Assumptions:

• First speed v1 = 90 km/h = 25 m/s; time t1 = 36 s.
• Second speed v2 = 45 km/h = 12.5 m/s; length L2 = L1 - 100 m.
• Same bridge; uniform speeds.

Concept / Approach:
For the first train: v1 * t1 = L1 + B, hence B = 25 * 36 - L1 = 900 - L1. For the second: time t2 satisfies v2 * t2 = L2 + B = (L1 - 100) + (900 - L1) = 800, eliminating L1.

Step-by-Step Solution:

Verification / Alternative check:
Whatever the actual L1 is, the combined expression simplifies to 800 m. The second train at 12.5 m/s must therefore take 64 s.

Why Other Options Are Wrong:
60 s, 72 s: Correspond to distances 750 m and 900 m, inconsistent with 800 m. “1 hr” is obviously not plausible for a bridge crossing.

Common Pitfalls:
Trying to find the bridge and first train lengths separately; unnecessary algebra causes errors. Use elimination as shown.

Final Answer:
64 sec