An elevator is moving upward 1.10 m/s when it experiences an acceleration of 0.39 m/s^2 downward, over a distance of 0.85 m. What will its final speed be? Answer in units of m/s
I need someone to verify and/or explain.
I got 0.580517 m/s.
Work:
v^2 = vo^2 + 2ax
v=final velocity
vo=initial velocity
a= acceleration
x=change in position
v^2=(1.0^2) + 2 * (-0.39) * (0.85) = 0.337
sqrt(0.337) = 0.580517 m/s
I wasn't sure if the distance should be -0.85 because going downward would make it negative?
I need someone to verify and/or explain.
I got 0.580517 m/s.
Work:
v^2 = vo^2 + 2ax
v=final velocity
vo=initial velocity
a= acceleration
x=change in position
v^2=(1.0^2) + 2 * (-0.39) * (0.85) = 0.337
sqrt(0.337) = 0.580517 m/s
I wasn't sure if the distance should be -0.85 because going downward would make it negative?
-
Your method is correct.
but the answer seems too low.
I think you used 1.0 m/s where you should have used 1.10 m/s
Once you decided that V= 1.10 m/s you have already decided that up is positive.
The elevator is NOT moving down. It is going up and slowing down.
So the positive for 0.85 m is correct.
It did move 0.85 m UPWARDS.
answer = 0.74 m/s upwards.
but the answer seems too low.
I think you used 1.0 m/s where you should have used 1.10 m/s
Once you decided that V= 1.10 m/s you have already decided that up is positive.
The elevator is NOT moving down. It is going up and slowing down.
So the positive for 0.85 m is correct.
It did move 0.85 m UPWARDS.
answer = 0.74 m/s upwards.