An elevator is moving upward at 1.20 m/s when it experiences an acceleration 0.25 m/s2 downward, over a distance of 0.71 m. What will its final velocity be?
Scroll down a bit to my answer to Dakota. Same problem, different numbers.
To find the final velocity of the elevator, we can use the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity
vi = initial velocity
a = acceleration
d = distance
Given:
Initial velocity (vi) = 1.20 m/s (upward direction)
Acceleration (a) = -0.25 m/s^2 (downward, negative because it's opposite to the initial velocity)
Distance (d) = 0.71 m
First, let's determine the initial velocity when the elevator reaches its final position after experiencing the given acceleration. We'll use a formula for calculating the final velocity when accelerating for a given distance:
vf = vi + at
Substituting the given values:
vf = 1.20 m/s + (-0.25 m/s^2)(0.71 m)
vf = 1.20 m/s - 0.1775 m/s
vf = 1.02 m/s
Now, we can substitute this value into the equation of motion to find the final velocity:
vf^2 = vi^2 + 2ad
(1.02 m/s)^2 = (1.20 m/s)^2 + 2(-0.25 m/s^2)(0.71 m)
1.0404 m^2/s^2 = 1.44 m^2/s^2 - 0.355 m^2/s^2
1.0404 m^2/s^2 = 1.085 m^2/s^2
√(1.0404 m^2/s^2) = √(1.085 m^2/s^2)
vf = 1.020 m/s
Therefore, the final velocity of the elevator will be approximately 1.020 m/s.