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Old Sep 13th 2016, 03:21 PM   #1
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Join Date: Sep 2016
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Smile Acceleration Help?!

Hi, guys! I really need your help with a problem in my homework.

An empty elevator is travelling upward at 1.25 m/s passing the 3rd floor when it suddenly receives a request from the 1st floor 9.50 m below. It immediately begins to accelerate at 0.310 m/s^2 downward. When it is 2.50 m from the 1st floor, it begins to accelerate to stop at the 1st floor.

a) What is the magnitude and direction of the elevator's final acceleration?

b) How long must the passenger wait from the time they push the button until the elevator arrives?

I would greatly appreciate any help anyone can offer. I've been working on it for a week now and I've still got nothing.
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Old Sep 16th 2016, 05:55 AM   #2
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Location: Morristown, NJ USA
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The elevator has two segments which you must consider separately. In segment 1 the elevator goes from y=9m to y=2.5m under constant acceleration. You know the initial and final positions and initial velocity. You don't know either the final velocity (when the elevator is 2.5 m above the ground) nor the time it takes to get there. You can use either of the standard equations of motion to get started:

1. You can use v2^2 - v1^2 = 2ad, which will give you the final velocity. From that you can use delta v = a delta t to calculate the time. Then figuring the acceleration and time for the 2nd segment is straightforward. The only "trick" here is to watch your signs - remember that the square root of a number may be either positive o negative, so be sure to use the correct one.

2. Method 2 is to use d2 = d1 + v1t + (1/2)at^2. You have everything you need to calcuklate the time (you'll neeed to solve a quadratic equation). Then calculate v2 from v2 = v1+at. The rest is easy.

My advice is to try both methods - it will help give you confidence in manipulating these equations.
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