Question

At a metro station, a girl walks up a stationary escalator in time t 1 . If she remains stationary on
the escalator, then the escalator take her up in time t 2 . The time taken by her to walk up on the
moving escalator will be then,
A. $\dfrac{{{t_1} + {t_2}}}{2}$
B. $\dfrac{{{t_1}{t_2}}}{{{t_2} - {t_1}}}$
C. $\dfrac{{{t_1}{t_2}}}{{{t_2} + {t_1}}}$
D. ${t_1} - {t_2}$

Answer 1

Hint: The equation of motion $s = vt$has to be applied to obtain the time taken by the girl to
walk up the escalator. The velocity of the girl and the escalator has to be assumed in each
condition.
Complete step by step solution:
Given,
Time taken by the girl to go up a stationary escalator be ${t_1}$
Time taken by the girl when she is stationary on the escalator be ${t_2}$
Let speed of the girl on stationary escalator be ${v_1}$
Let speed of the escalator be ${v_2}$
Let the distance be $d$.
The time taken to go up when the escalator is stationary,

${t_1} = \dfrac{d}{{{v_1}}}$ …….(1)
The velocity of the girl,
${v_1} = \dfrac{d}{{{t_1}}}$ …….(2)
The velocity of the escalator,
${v_2} = \dfrac{d}{{{t_2}}}$ …….(3)
Now, time taken when both the girl and escalator moves up is,
$t = \dfrac{d}{{{v_1} + {v_2}}}$ …….(4)
Now, substituting values of ${v_1}$and ${v_2}$from equation (2) and equation (3) in equation
(1) we get,
$\begin{array}{l}t = \dfrac{d}{{\dfrac{d}{{{t_1}}} + \dfrac{d}{{{t_2}}}}}\\ \Rightarrow t =
\dfrac{{{t_1}{t_2}}}{{{t_2} + {t_1}}}\end{array}$
Hence the time taken by the girl to walk up on the moving escalator is
$\dfrac{{{t_1}{t_2}}}{{{t_2} + {t_1}}}$
Hence the correct answer is (C).
Note: The students have to apply the equation of motion concept. The individual time taken by
the girl at stationary and moving escalator has to be found in terms of distance and time since
time is distance/velocity. The total time can be calculated from the individual times.