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Question:

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her walk up on the moving escalator will be:

t1t2/(t2+t1)

t1+t2/2

t1−t2

t1t2/(t2−t1)

Solution:

t1=time at stationary
t2=time by escalator
displacement of escalator
v1=Velocity of stationary
v2=Velocity of Preeti
(V1+V2)=d/t′, as
V1=d/t1,V2=d/t2
d/t1+d/t2=d/t′
→t′=t1t2/(t1+t2)