Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \( t_1 \). On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \( t_2 \). The time taken by her to walk up on the moving escalator will be:
(2017-Delhi)
\( \frac{t_1 t_2}{t_2 - t_1} \)
\( \frac{t_1 t_2}{t_2 + t_1} \)
\( \frac{t_1 + t_2}{2} \)
Solution:
Concept: Relative velocity.
If Preeti's speed is \( v_p \) and escalator's speed is \( v_e \), for total length \( L \), \( v_p = L/t_1 \) and \( v_e = L/t_2 \). When Preeti walks on moving escalator, effective speed is \( v_p + v_e \). Time taken \( T = L / (v_p + v_e) = L / (L/t_1 + L/t_2) = \frac{t_1 t_2}{t_1 + t_2} \).
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