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 1.80 m/s21.80 m/s2 or travel faster than 16.8 m/s16.8 m/s, what is the minimum time in which an elevator can travel the 373373 m from the ground floor to the observatory floor?

minimum travel time: ? s

Answer 1

As per Law ,Max allowable velocity = 16.8m/s

Max acceleration allowed = 1.8m/s²

Total height to cover = 373 m

The journey can be divide into three: Firstly the elevator accelerate from rest,reach max velocity.

2.Then travel with constant velocity

3. finally decelerate coming to stop.

First stage

initial velocity =0 m/s ,final velocity =16.8m/s and acceleration =1.8m/s²

Find time and distance covered

Use Formula $v=u+at$

$16.8m/s=0m/s+1.8m/s^{2}*t_{1}$

$16.8=0+1.8*t_{1}$

${\color{Red} t_{1}=9.33s}$

-------------

Use Formula $v^{2}-u^{2}=2as$

$(16.8m/s)^{2}-(0m/s)^{2}=2*1.8m/s^{2}*s_{1}$

$16.8^{2}-0^{2}=2*1.8*s_{1}$

$16.8^{2}=3.6s_{1}$

${\color{Red} s_{1}=78.4m}$

================

Final stage (decelerating)

Final velocity =0m/s ,initial velocity =16.8m/s , deceleration =1.8m/s²

Use Formula $v=u+at$

$0m/s=16.8m/s-1.8m/s^{2}*t_{3}$

$0=16.8-1.8*t_{3}$

${\color{Red} t_{3}=9.33s}$

--------------

Use Formula $v^{2}-u^{2}=2as$

$(0m/s)^{2}-(16.8m/s)^{2}=-2*1.8m/s^{2}*s_{3}$

$0^{2}-16.8^{2}=-2*1.8*s_{3}$

$16.8^{2}=3.6s_{3}$

${\color{Red} s_{3}=78.4m}$

===============

The Middle Part of Journey

Constant velocity =16.8m/s

distance to cover = $373m-s_{1}-s_{3}=373m-78.4m-78.4m=216.2m$

${\color{Red} s_{2}=216.2m}$

Speed =distance /time

time =distance /speed

$t_{2}=s_{2}/16.8m/s$

$t_{2}=216.2m/16.8m/s$

${\color{Red} t_{2}=12.869s}$

===========

Total time ,$t=t_{1}+t_{2}+t_{3}$

$t=9.33s+12.869s+9.33s$

ANSWER: ${\color{Red} t=31.529s}$

=====================