Motion in a Straight Line - Result Question 2
2. 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:
[2017]
(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) $\frac{t_1+t_2}{2}$
Show Answer
Answer:
Correct Answer: 2. (b)
Solution:
- (b) Let the distance be ’ $d$ ’ time taken by preeti to travel up the stationary escalator $=t_1$ Velocity of preeti w.r.t. elevator $v_1=\frac{d}{t_1}$
Since the distance is same let the time taken when preeti stands on the moving escalator $=t_2$.
Velocity of elevator w.r.t. ground $v_2=\frac{d}{t_2}$
Then net velocity of preeti w.r.t. ground
$ \begin{aligned} & v=v_1+v_2 \\ & \frac{d}{t}=\frac{d}{t_1}+\frac{d}{t_2} \\ & \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2} \end{aligned} $
$\therefore t=\frac{t_1 t_2}{(t_1+t_2)}$ (time taken by preeti to walk
up on the moving escalator)
