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?
.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
To find the final velocity of the elevator, we can use the formula:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Given:
Initial velocity, u = 0.90 m/s (upward)
Acceleration, a = -0.40 m/s^2 (downward, negative sign indicates downward direction)
Distance, s = 0.64 m
Let's substitute these values into the equation:
v^2 = (0.90 m/s)^2 + 2(-0.40 m/s^2)(0.64 m)
Simplifying this equation:
v^2 = 0.81 m^2/s^2 - 0.512 m^2/s^2
v^2 = 0.298 m^2/s^2
To find the final velocity, take the square root of both sides:
v = sqrt(0.298 m^2/s^2)
v ≈ 0.546 m/s
Therefore, the final velocity of the elevator will be approximately 0.546 m/s.