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 3.30 m/s2 or travel faster than 19.3 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

Assuming that the law applies also while decelerating,that is while stopping, a= -3.3 m/s² . Also time taken(t) during acceleration and decelration is same (=t, say)

let total time be T

total distance = 2*(distance travelled in acceleration) + (distance travelled at constant velocity)

373 = 2*s + v*(T-2t) -----------(1)

distance travelled in acceleration:

u=0; v=19.3 m/s; a=3.3 m/s

t=(v-u)/a = 19.3/3.3 = 5.848 seconds

now using: v² - u²= 2as

s= 19.3²/2*3.3 = 56.437 m

Putting the values of s and t in equation (1)

373 = 2*56.437 + 19.3*(T-2*5.848)

T= 25.174 seconds