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A person walks up a stalled escalator in...

A person walks up a stalled escalator in 90 s. When standingon the same escalator, now moving, he is carried in 60 s.The time it would take him to walk up the moving escalator will be:

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Let the length of escalator be L. if `v` is the velocity of man (relative to escalator) and `V` of escalator, then according to given problem
`(L)/(v) = 90 s`....(i) and `(L)/(V) = 60 s`....(ii)
Now if the person walks up on the moving escalator his velocity relative to the ground will be `(V + v)`. So time taken by him to move a distance L relative to the ground will be,
`t = (L)/((V + v))` or `(1)/(t) = (V)/(L) + (v)/(L)`
which in the light of Eqns. (i) and (ii) gives
`(1)/(t) = (1)/(60) + (1)/(90)`, i.e., `t = 36 s`
i.e., the person will take 36 s to walk up to the moving escalator.
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