Question

Ruma reached the metro station and found that the escalator was not working. She walked up the stationary escalator with velocity \( v_1 \) in time \( t_1 \). On another day, if she remains stationary on the escalator moving with velocity \( v_2 \), the escalator takes her up in time \( t_2 \). The time taken by her to walk up with velocity \( v_1 \) on the moving escalator will be:

• A. \( \frac{t_1}{t_2} \)
• B. \( \frac{t_1 + t_2}{t_2 - t_1} \)
• C. \( \frac{t_1 + t_2}{v_1 + v_2} \)
• D. \( \frac{t_1 t_2}{t_1 + t_2} \)

Hint:

When dealing with combined motion, remember to add the velocities when both are in the same direction. The time taken will be inversely proportional to the sum of the velocities.