A passenger runs down the descending escalator at a speed υ = 2 m / s relative to the escalator.

| February 26, 2021 | education

A passenger runs down the descending escalator at a speed υ = 2 m / s relative to the escalator. The escalator speed is u = 1 m / s. The number of escalator steps on the descent is N = 90. How many steps will the passenger go down the escalator?

the solution is given N = 90 – the number of steps on the escalator v = 2 m / s – the speed of the passenger running down u = 1 m / s – the speed of the escalator then we find the total speed u + v relative to the ground х – “step length” N1 – the number of steps which a passenger will pass in time t passes the path S1 = N * x = (u + v) * t relative to the ground in time t passes the path S2 = N1 * x = (v) * t relative to the escalator the proportion N * x / N1 * x = ( u + v) * t / (v) * t N / N1 = (u + v) / (v) N1 = N * v / (u + v) = 9 0 * 2 / (1 + 2) = 60
answer 60



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