Question

You 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 more than 4.50 m/s24.50 m/s2 or travel faster than 15.3 m/s15.3 m/s, what is the minimum time in which an elevator can travel the 373373 m from the ground floor to the observatory floor?

Answer 1

The elevator will start from rest, accelerate with the maximum acceleration, 4.5 m/s², reach the maximum velocity of 15.3 m/s. Finally, it will decelerate with 4.5 m/s² before coming to rest on the observatory floor.

Time taken and the distance travelled during the acceleration and deceleration part is the same.
Consider the acceleration part,
Initial velocity, u = 0
Final velocity, v = 15.3 m/s
Acceleration, a = 4.5 m/s²

Using the formula, t = (v - u) / a,
t = (15.3 - 0) / 4.5
= 3.4 s
Using the formula, the distance travelled, s = u * t + 0.5 * a * t²,
s = 0 * 3.4 + 0.5 * 4.5 * 3.4²
= 26.01 m
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Consider the part where the accelerator is moving at constant velocity of 15.3 m/s
Total distance travelled during the constant velocity part, d = 373 - 2 * s
d = 373 - 2 * 26.01
= 320.98 m
Time taken to complete the constant velocity part of the journey,
t' = d / v
= 320.98 / 15.3
= 20.98 s
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Total time take for the whole journey = Time taken during the acceelration + time taken during the constant velocity part + time taken during the deceleration
= t + t' + t
= 3.4 + 20.98 + 3.4
= 27.8 s