Q: At a metro station, a girl walks up a stationary escalator in time t1 . If she remains stationary on the escalator, then the escalator take her up in time t2 . The time taken by her to walk up on the moving escalator will be
(a) (t1 + t2 )/2
(b) t1 t2/(t2-t1 )
(c) t1 t2/(t2 + t1 )
(d) t1 – t2
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Ans: (c)
Sol: Let v1 be the velocity of girl on stationary escalator and d is the distance travelled by her
Hence , $v_1 = \frac{d}{t_1}$
Let v2 be the velocity of escalator
$v_2 = \frac{d}{t_2}$
Effective velocity of girl on moving escalator w.r.t ground is
$v = v_1 + v_2 $
$v = \frac{d}{t_1} + \frac{d}{t_2} = d (\frac{t_1 + t_2}{t_1 \; t_2})$
Time taken by girl to walk up on moving escalator will be
$ t = \frac{d}{v} = \frac{d}{d (\frac{t_1 + t_2}{t_1 \; t_2})}$
$ t = \frac{t_1 \; t_2}{t_1 + t_2 }$