An elevator is moving upward 1.17 m/s when
it experiences an acceleration of 0.33 m/s2
downward, over a distance of 0.75 m.
What will its final speed be?
Answer in units of m/s
To find the elevator's final speed, we need to use the equation for motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
In this case, the elevator's initial velocity (u) is 1.17 m/s upward, the acceleration (a) is 0.33 m/s^2 downward, and the distance (s) is 0.75 m.
First, we need to determine the direction of the acceleration. Since the elevator is moving upward and experiences a downward acceleration, we can assume the final velocity will be downward.
Now, let's substitute the given values into the equation:
v^2 = (1.17^2) + 2*(-0.33)*(0.75)
Simplifying the equation:
v^2 = 1.3689 - 0.495
v^2 = 0.8739
To find the final velocity (v), we take the square root of both sides of the equation:
v = √0.8739
Using a calculator or any math software, we find:
v ≈ 0.934 m/s
Therefore, the elevator's final speed will be approximately 0.934 m/s downward.