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.

Answer 1

The Correct Option is C

Step 1: Problem Setup 1. Stationary escalator: Ruma's velocity is \( v_1 \), time is \( t_1 \), and height \( h \) is: \[ h = v_1 \cdot t_1 \] 2. Moving escalator (Ruma stationary): Escalator velocity is \( v_2 \), time is \( t_2 \), and height \( h \) is: \[ h = v_2 \cdot t_2 \] Step 2: Ruma Walks Up Moving Escalator Ruma's effective velocity is \( v_1 + v_2 \), and time \( t_3 \) to cover height \( h \) is: \[ t_3 = \frac{h}{v_1 + v_2} \] Step 3: Substituting for \( h \) Substitute \( h \) from the stationary and moving escalator cases into the \( t_3 \) equation: \[ t_3 = \frac{v_1 \cdot t_1}{v_1 + v_2} \] Step 4: Conclusion Time for Ruma to walk up the moving escalator with velocity \( v_1 \) is: \[ t_3 = \frac{t_1 \cdot v_1}{v_1 + v_2} \] Comparing this with the options, the correct answer is Option (C). Final Answer: The correct answer is: \[ \boxed{(C)} \frac{t_1 + t_2}{v_1 + v_2} \]