Question

Preeti reached a 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 up on a moving escalator will be

• A. $\frac{t_1{t}_2}{t_2-t_1}$
• B. $\frac{t_1{t}_2}{t_2+t_1}$
• C. $t_1-t_2$
• D. $\frac{t_1+t_2}{2}$

Answer 1

The correct option is B $\frac{t_1{t}_2}{t_2+t_1}$

Let $v_1$ be the velocity of Preeti and $v_2$ be the velocity of of the escalator and $l$ be the distance moved.

When she remains stationary on the moving escalator, the escalator takes her up in time $t_2$,

$t_2=\frac{l}{v_2}.....\left(i\right)$

When she moves along with the escalator, her velocity w.r.t the ground will be $v_1+v_2$

$\Rightarrow t=\frac{l}{v_1+v_2}.....\left(ii\right)$

where $t\to$ time taken to travel the distance $l$ when both the escalator and Preeti are moving.

When the escalator is stationary, she walks up the stationary escalator in time $t_1$

$t_1=\frac{l}{v_1}.....\left(iii\right)$

From equations $\left(i\right)$, $\left(ii\right)$ & $\left(iii\right)$,

$t=\frac{l}{v_1+v_2}$

$\Rightarrow t=\frac{l}{\frac{l}{t_1}+\frac{l}{t_2}}$

$\therefore t=\frac{t_1{t}_2}{t_2+t_1}$