An elevator is moving upward at 1.00 m/s when it experiences an acceleration 0.28 m/s2 downward, over a distance of 0.74 m. What will its final velocity be?
To find the final velocity of the elevator, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the distance
Given:
- Initial velocity, u = 1.00 m/s (upward)
- Acceleration, a = -0.28 m/s^2 (downward, negative sign indicates the opposite direction of the initial velocity)
- Distance, s = 0.74 m
Substituting these values into the equation of motion:
v^2 = (1.00 m/s)^2 + 2(-0.28 m/s^2)(0.74 m)
Simplifying:
v^2 = 1.00 m^2/s^2 - 0.41 m^2/s^2
v^2 = 0.59 m^2/s^2
Taking the square root of both sides to find v:
v = √(0.59 m^2/s^2)
Calculating the square root:
v ≈ 0.77 m/s
Therefore, the final velocity of the elevator will be approximately 0.77 m/s.