Q: 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

(a) (t_{1} + t_{2} )/2

(b) t_{1} t_{2}/(t_{2}-t_{1} )

(c) t_{1} t_{2}/(t_{2} + t_{1} )

(d) t_{1} – t_{2}

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_{1}be the velocity of girl on stationary escalator and d is the distance travelled by her

Hence , $v_1 = \frac{d}{t_1}$

Let v_{2} 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 }$