Question

An elevator accelerates from rest to a velocity of 12.0 m/sec as it rises 28.0 meters.  What was its’ acceleration, and for what time did it travel while it was accelerating?​

Answer 1

given that

V = 12.0 m/s

U = 0 m/s

s = 28 m

Step-by-step explanation:

using the equation of motion

V^2 = U^2 + 2as

12^2 = 0 + 2(28)a

144 = 56a

a = 5.5 m/s^2

v = u +at

12 = 0 + 5.5t

12 = 5.5t

t = 2.2s