An elevator is moving upward at 0.90 m/s when it experiences an acceleration 0.40 m/s2 downward, over a distance of 0.64 m. What will its final velocity be?

Answer 1

.64 = .9 t - .5(.4)t^2
.2 t^2 - .9 t + .64 = 0
t = [.9 +/- sqrt (.89 - .512) ]/.4
t = [ .9 +/- .615 ]/ .4
we want on the way up so use shorter time
t = .713 seconds
v = Vi + a t
v = .9 - .4 (.713)
v = .615 m/s