Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time . On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time . The time taken by h

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Solution

Q.
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]

1 $\frac{t_{1} t_{2}}{t_{2} - t_{1}}$

2 $\frac{t_{1} t_{2}}{t_{2} + t_{1}}$

3 $t_{1} - t_{2}$

4 $\frac{t_{1} + t_{2}}{2}$

Ans.
(2)

Let $v_{1}$ is the velocity of Preeti on stationary escalator and $d$ is the distance travelled by her $∴$ $v_{1} = \frac{d}{t_{1}}$

Again, let $v_{2}$ be the velocity of escalator $∴$ $v_{2} = \frac{d}{t_{2}}$

$∴$ Net velocity of Preeti on moving escalator with respect to the ground

$v = v_{1} + v_{2} = \frac{d}{t_{1}} + \frac{d}{t_{2}} = d (\frac{t_{1} + t_{2}}{t_{1} t_{2}})$

The time taken by her to walk up on the moving escalator will be

$t = \frac{d}{v} = \frac{d}{d (\frac{t_{1} + t_{2}}{t_{1} t_{2}})} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

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