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 \( v_2 \), then 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. \( \frac{t_1t_2}{t_2-t_1} \)
• B. \( \frac{t_1t_2}{t_1+t_2} \)
• C. \( \frac{t_1-t_2}{t_1+t_2} \)
• D. \( \frac{t_1+t_2}{2(t_1-t_2)} \)

Hint:

Relative speed problems: \begin{itemize} \item Add speeds when moving together. \item Use common length. \end{itemize}

Answer 1

The Correct Option is B

Concept: Let escalator length = \( L \). Step 1: {\color{red}Stationary escalator.} \[ v_1=\frac{L}{t_1} \] Step 2: {\color{red}Standing on moving escalator.} \[ v_2=\frac{L}{t_2} \] Step 3: {\color{red}Both moving.} Net speed: \[ v=v_1+v_2 \] Time: \[ t=\frac{L}{v_1+v_2} \] Step 4: {\color{red}Substitute.} \[ t=\frac{L}{\frac{L}{t_1}+\frac{L}{t_2}} =\frac{1}{\frac{1}{t_1}+\frac{1}{t_2}} \] \[ t=\frac{t_1t_2}{t_1+t_2} \]