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At a metro station $,$ a girl walks up a stationary escalator in time $6 s .$ If she remains stationary on the escalator $,$ then the escalator take her up in time $3 s .$ The time taken by her to walk up on the moving escalator will be ……. sec.

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answer is 2.

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Detailed Solution

Let $L$ be the length of escalator $.$

Velocity of girl $w . r . t .$ escalator $, v_{g e} = \frac{L}{t_{1}}$

Velocity of escalator $, v_{e} = \frac{L}{t_{2}}$

Velocity of girl w.r.t. ground would be

$v_{g} = v_{g e} + v_{e} = L (\frac{1}{t_{1}} + \frac{1}{t_{2}})$

The desired time is $t = \frac{L}{v_{g}} = \frac{L}{L [\frac{1}{t_{1}} + \frac{1}{t_{2}}]} = \frac{t_{1} t_{2}}{t_{1} + t_{2}} = \frac{6 \times 3}{6 + 3} = 2 \sec$

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