At a metro station, a girl walks up a stationary escalator in time \(t_1\). If she remains stationary on the escalator, then the escalator takes her up in time \(t_2\). The time taken by her to walk upon the moving escalator will be:
1. \( \left(\mathrm{t}_1+\mathrm{t}_2\right) / 2\)
2. \( \mathrm{t}_1 \mathrm{t}_2 /\left(\mathrm{t}_2-\mathrm{t}_1\right)\)
3. \( \mathrm{t}_1 \mathrm{t}_2 /\left(\mathrm{t}_1+\mathrm{t}_2\right) \)
4. \( \mathrm{t}_1-\mathrm{t}_2\)

(c) Hint: Use the concept of relative motion.

Step 1: Find the velocity of the girl w.r.to the ground.

In this question, we have to find the net velocity with respect to the ground that will be equal to the velocity of the girl plus the velocity of the escalator.

Let displacement is L, then,

 velocity of girl, vg=Lt1 velocity of escalator, ve=Lt2 Net velocity of the girl =vg+ve=Lt1+Lt2

Step 2: Find the time taken by the girl.

If t is the total time taken by the girl in covering distance L, then,

Lt=Lt1+Lt2t=t1t2t1+t2