Question

Let the length of escalator is L. Then, velocity of man = distance/time taken ( =mathrm{L} / mathrm{T}_{1} ) Velocity of man with respect to escalator ( =mathrm{L} / mathrm{T}_{2} )
so, velocity of escalator with respect to man ( =-mathrm{L} / mathrm{T}_{2} )
We know, velocity of escalator = velocity of escalator with respect to man
+ velocity of man ( mathrm{L} / mathrm{T}=-mathrm{L} / mathrm{T}_{2}+mathrm{L} / mathrm{T}_{1}=mathrm{L} /left(1 / mathrm{T}_{1}-1 / mathrm{T}_{2}right) )
( mathrm{T}=mathrm{T}_{1} mathrm{T}_{2} /left(mathrm{T}_{2}-mathrm{T}_{1}right) )

# (4) 5 m/s 22. A person takes T, second on the stationary escalator to cover some distance. The person takes T, second in the moving escalator to cover same distance by same velocity. If person is at rest on the escalator then time taken by him to cover same distance is (1) TORT T -T, (4) Tz +T

Solution