u are an engineer in charge of designing a new generation of elevators for a pro
ID: 2225290 • Letter: U
Question
u are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators. If state law mandates that elevators cannot accelerate at greater than 2.10 m/s2 or travel faster than 14.8 m/s, what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor?Explanation / Answer
Assuming that the law also applies to the deceleration when slowing to a stop .. max deceleration will be -5.10 m/s² .. and the periods (t) of max acceleration and deceleration will be the same.
Total distance = 2(accel distance) + distance at max speed
Let t = accelerating or decelerating time, and total time = T
373m = 2(av vel x t) + (v x [T - 2t])
373 = vt + v[T - 2t]
373 = v(t + T - 2t)
373 = v(T - t) as v(max) = 10.30m/s
373 = 10.3(T-t) ---- (1)
From a = v/t .. t = v/a = 10.30m/s / 5.10 => t = 2.01 s ---- (2)
Sub 2 into 1 ..
373 = 10.3(T - 2.01) => T = (373/10.30) + 2.01 .. .. .. T = 38.22s
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