Preeti reached the metro station and found that the escalator was not working. She walked up the sta

Question

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 down on the upward moving escalator will be (Assuming her walking speed is greater than the escalator's speed)

Professor PC Thomas Classes · Answer

(a) : Let $v_{1}$ be the velocity of Preeti on a stationary escalator and $d$ be the distance $	herefore v_{1}=\frac{d}{t_{1}}$ \ Let $v_{2}$ be the velocity of the upward escalator $	herefore v_{2}=\frac{d}{t_{2}}$ \ Net velocity of Preeti walking down the upward moving escalator with respect to the ground: \ $v_{	ext{net}} = v_{1} - v_{2} = \frac{d}{t_{1}} - \frac{d}{t_{2}} = d\left(\frac{t_{2}-t_{1}}{t_{1} t_{2}}ight)$ \ The time taken by her to walk down is $t = \frac{d}{v_{	ext{net}}} = \frac{d}{d\left(\frac{t_{2}-t_{1}}{t_{1} t_{2}}ight)} = \frac{t_{1} t_{2}}{t_{2}-t_{1}}$