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 at greater than 4.50 m/s2 or travel faster than 19.8 m/s, what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor?

Answer 1

23.24 seconds.

The travel time from the start of the trip to the end will have three phases. Initial acceleration, cruising, and deceleration. So, let's see how long each phase takes. For the initial acceleration phase, assume it accelerates at 4.50m/s2 until it reaches 19.8 m/s. So calculate t. t = 19.8 m/s / 4.5 m/s2 = 19.8/4.5 s = 4.4 s

The distance traveled during the acceleration phase will be 1/2AT^2. So d = 0.5 * 4.5 * t^2 = 2.25 * 4.4^2 = 2.25 * 19.36 = 43.56 m

The deceleration phase will also take 4.4 seconds and occur over a distance of 43.56 m. So the combined distance for acceleration and deceleration is 43.56 * 2 = 87.12 meters. Now the total distance that we want to travel is 373 m. So you can determine the distance for the cruising phase by subtracting the combined distances for acceleration and deceleration. Giving cruise = 373 - 87.12 = 285.88 m

The time the cruising phase takes is the distance divided by the speed, so we get t = 285.88 m / 19.8 m/s = 14.44 seconds.

Finally, the total travel time is the time for acceleration plus time for cruising plus time for deceleration, giving 4.4 s + 14.44 s + 4.4 s = 23.24 seconds.