Question

An elevator is designed to travel to the top of building with 35 floors. The elevator has a top speed of 4.00 m/s and a constant acceleration of 3.00 m/s^2 when accelerating from rest to its top speed, and from its top speed back to rest. Assume that the height of each floor is 4.30 m. How long does it take the elevator to reach the top floor?

Answer 1

39.0 s

Step-by-step explanation:

First find the time and distance the elevator takes to accelerate or decelerate.

Given:

u = 0 m/s

v = 4.00 m/s

a = 3.00 m/s²

Find: t

v = at + u

4.00 = (3.00) t + 0

t = 1.333 s

Find: s

v² = v₀² + 2as

(4.00)² = (0)² + 2(3.00)s

s = 2.667 m

Next, find the time the elevator spends at constant speed.

Given:

s = 35 × 4.30 m − 2 × 2.667 m = 145.167 m

u = 4.00 m/s

a = 0 m/s²

Find: t

s = u t + ½ a t²

145.167 = (4.00) t + ½ (0) t²

t = 36.3 s

So the total time is:

2 (1.333 s) + 36.3 s = 39.0 s