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Acceleration/distance/time problem.. |
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| Sep10-07, 06:10 AM | #1 |
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Acceleration/distance/time problem..
For our investigatory project, I and my groupmates are going to test the efficiency of water cress in cleaning wastewater. We're going to adapt the concept of the Green Roof Water Recycling system, where the wastewater from kitchen and bathroom sinks gets pumped up to the roof and gets trickled onto some semi-aquatic plants. Gravity brings the water back down for reuse. Anyway, I thought I might figure out how long I should put the water cress in the wastewater using the acceleration of gravity (9.8 m/s2) and the length/distance of the original GROW set-up, which is 4m. The question is, what's the equation going to be? Something similar to, um, the time equation involving speed (t = d/s)?
I really hope someone can answer these questions for me... thank you! ^___^ |
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| Sep10-07, 06:25 AM | #2 |
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I'm not too sure what exactly you are asking? What is your aim? ie You say you need the right length, but right length for what?
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| Sep10-07, 07:49 AM | #3 |
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I'm sorry. I meant what's the time, given that the acceleration is 9.8 m/s2 and the distance is 4m? :)
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| Sep10-07, 07:57 AM | #4 |
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Acceleration/distance/time problem..
The question you are asking is: How long will an object take to fall 4m, with acceleration at 9.8 m/s^2?
We use a handy equation here: [tex] s = ut + \frac{1}{2} at^2[/tex] where s is the displacement, u in the initial velocity, a is acceleration and t is time. We know s=4, u is 0 since when it starts falling its originally not moving. So the equation then becomes [tex]4 = \frac{1}{2}(9.8)t^2[/tex]. Times both sides by 2, then divide both by 9.8. [tex]\frac{8}{9.8} = t^2[/tex]. Now just square root the stuff on the left hand side and thats how long it takes :) My estimate is about 0.9 seconds, but I have no calculator. |
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