An elevator is moving upward at 1.03 m/s when it experiences an acceleration 0.31 m/s2 downward, over a distance of 0.73 m. What will its final velocity be?

v₀²-v² = 2as

v=sqrt(v₀²-2as)

Can somebody help me solve this?

To find the final velocity of the elevator, we need to use the equation of motion that relates initial velocity (v₀), final velocity (v), acceleration (a), and displacement (d):

v² = v₀² + 2ad

We are given:
- Initial velocity (v₀) = 1.03 m/s (upward)
- Acceleration (a) = 0.31 m/s² (downward)
- Displacement (d) = 0.73 m

Let's plug in the values into the equation and solve for the final velocity (v):

v² = (1.03 m/s)² + 2(0.31 m/s²)(0.73 m)

First, square the initial velocity:
v² = 1.06 m²/s² + 2(0.31 m/s²)(0.73 m)

Next, calculate the product of 2(0.31 m/s²)(0.73 m):
v² = 1.06 m²/s² + 0.4526 m²/s²

Add the two terms together:
v² = 1.5126 m²/s²

Finally, take the square root of both sides to find the final velocity (v):
v = √(1.5126 m²/s²)

Using a calculator, we find that the square root of 1.5126 is approximately 1.23 m/s.

Therefore, the final velocity of the elevator will be approximately 1.23 m/s.