Question

If state law mandates that elevators cannot accelerate more than 4.80 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

Answer:

23.0 s

Explanation:

Given:

v₀ = 0 m/s

v = 19.8 m/s

a = 4.80 m/s²

Find: Δx and t

v² = v₀² + 2aΔx

(19.8 m/s)² = (0 m/s)² + 2 (4.80 m/s²) Δx

Δx = 40.84 m

v = at + v₀

19.8 m/s = (4.80 m/s²) t + 0 m/s

t = 4.125 s

The elevator takes 40.84 m and 4.125 s to accelerate, and therefore also 40.84 m and 4.125 s to decelerate.

That leaves 291.3 m to travel at top speed.  The time it takes is:

291.3 m / (19.8 m/s) = 14.71 s

The total time is 4.125 s + 14.71 s + 4.125 s = 23.0 s.