A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t₁. If he remains stationary on a moving escalator then the escalator takes him up in time t₂. The time taken by him to walk up on the moving escalator will be :

Solution

Physics sir · 3 steps

Do not add the times, add the speeds: each speed is just the escalator length over its own time.

1. Write both speeds. Let the escalator length be L. The boy's speed relative to the steps is v_b = L/t₁, and the escalator's speed is vₑ = L/t₂.
2. Speeds add. When he walks on the moving escalator, both carry him in the same direction. His speed relative to the ground is v_b + vₑ.

   t = L/(L/t₁ + L/t₂) = (t₁ t₂)/(t₁ + t₂)

3. So. L cancels, and the time is (t₁ t₂)/(t₁ + t₂), which is option (c). It is shorter than both t₁ and t₂, as it should be.

The idea to keep. When two motions push you along the same path, add their speeds, not their times. The reciprocals of the times add: 1/t = 1/t₁ + 1/t₂.

Answer from the NTA final answer key. Solution by PrepWiser.