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 2.40 m/s2 or travel faster than 18.8 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

Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

         v = v₀ + a t₁

the elevator starts with zero speed

         v = a t₁

         t₁ = v / a

         t₁ = 18.8 / 2.40

         t₁ = 7.833 s

in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

         y₁ = ½ a t₁²

         y₁ = ½ 2.40 7.833²

         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s