A person climbs up a stalled escalator in 60 s. If standingon the same but escalator running with constant velocityhe takes 40 s. How much time is taken by the person towalk up the moving escalator
Question
A person climbs up a stalled escalator in 60 s. If standingon the same but escalator running with constant velocityhe takes 40 s. How much time is taken by the person towalk up the moving escalator
Solution
Let's denote:
- T1 = time taken by the person to climb the stalled escalator = 60s
- T2 = time taken by the person to ride the moving escalator = 40s
- T3 = time taken by the person to walk up the moving escalator, which we want to find.
We can think of the person's speed and the escalator's speed as contributing to the total speed when the escalator is moving.
When the person climbs the stalled escalator, the speed is the person's walking speed (W). So, the distance (D) of the escalator is W*T1.
When the person rides the moving escalator, the speed is the escalator's speed (E). So, the same distance of the escalator is E*T2.
Therefore, we can write D = WT1 = ET2. From this, we can express the escalator's speed in terms of the person's walking speed and the times: E = W*T1/T2.
When the person walks up the moving escalator, the total speed is the sum of the person's speed and the escalator's speed (W + E). The time taken to travel the distance of the escalator at this speed is T3 = D/(W + E).
Substituting D = WT1 and E = WT1/T2 into this equation gives T3 = T1/(1 + T1/T2).
Substituting the given values T1 = 60s and T2 = 40s gives T3 = 60s/(1 + 60s/40s) = 60s/(1 + 1.5) = 60s/2.5 = 24s.
So, the person takes 24 seconds to walk up the moving escalator.
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