An elevator is moving upward at 0.93 m/s when it experiences an acceleration 0.29 m/s2 downward, over a distance of 0.67 m. What will its final velocity be?

h=(v²-vₒ²)/2a,

v=sqrt(2ah+vₒ²) = ...

To find the final velocity of the elevator, we can use the equation for uniformly accelerated motion:

vf^2 = vi^2 + 2ad

Where:
vf is the final velocity
vi is the initial velocity
a is the acceleration
d is the distance traveled

Given:
vi (initial velocity) = 0.93 m/s (upward)
a (acceleration) = -0.29 m/s^2 (downward)
d (distance) = 0.67 m

First, we need to determine the initial velocity in the downward direction. Since the acceleration is pointing downward, the initial velocity will also be in the downward direction. Therefore, the initial velocity is -0.93 m/s.

Now, we can substitute the known values into the equation:

vf^2 = (-0.93)^2 + 2(-0.29)(0.67)

vf^2 = 0.8649 - 0.3926

vf^2 = 0.4723

Taking the square root of both sides, we get:

vf = √0.4723

vf ≈ 0.687 m/s

Therefore, the final velocity of the elevator will be approximately 0.687 m/s.