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 up on the 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 velocity of preeti and $V_2$ be velocity of escalator and ${}^{\prime }{l}^{\prime }$ be the distance.
When she remains stationary on escalator,
$t_2=\frac{l}{V_2}.......\left(i\right)$
When she moves along with escalator, her velocity w.r.t ground will be, $(V_1+V_2)$
$\Rightarrow t=\frac{l}{V_1+V_2}......\left(ii\right)$
$t\to$ time to travel the distance ${}^{\prime }{l}^{\prime }$ when both escalator and person both moving.
When escalator is stationary,
$t_1=\frac{l}{V_1}.......\left(iii\right)$
From $Eq$. (i), (ii) & (iii)
$t=\frac{l}{V_1+V_2}$
$\Rightarrow t=\frac{l}{\frac{l}{t_1}+\frac{l}{t_2}}$
$\therefore t=\frac{t_1{t}_2}{t_2+t_1}$