Physics Help Forum Basic dynamics problem help

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 Feb 10th 2011, 12:34 AM #1 Junior Member   Join Date: Feb 2011 Posts: 3 Basic dynamics problem help The acceleration of a part undergoing a machining operation is measured and is determined to be a=12-6t mm/s^2. When t=0, v=0. For the interval of time from t=0 to t=4s, determine (a) the maximum velocity; (b) the displacement. where to begin on finding max velocity do i plug in 4 for t in the accelartion and solve for velocity. thanks in advanced Last edited by cramjm11; Feb 10th 2011 at 12:42 AM.
Feb 10th 2011, 04:32 AM   #2

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 Originally Posted by cramjm11 The acceleration of a part undergoing a machining operation is measured and is determined to be a=12-6t mm/s^2. When t=0, v=0. For the interval of time from t=0 to t=4s, determine (a) the maximum velocity; (b) the displacement. where to begin on finding max velocity do i plug in 4 for t in the accelartion and solve for velocity. thanks in advanced
You know that
a = 12 - 6t

You also know that
v(t) = (integral)(a) dt

So find v and then use Calculus to find where the maximum is. (This is essentially going to be the same as finding where a = 0. But make sure you understand why you need to do this.)

-Dan
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 Feb 15th 2011, 01:33 AM #3 Junior Member   Join Date: Feb 2011 Posts: 3 would max acceleration be at the max time of 4 sec which would be v-6t^2+12t 4 sec works out to 48mm/s2 at 2 sec velocity=0 so its paused plugging in and making a chart at every .5 sec it comes out to make a graph where it increases up to 6mm/s^2 at 1 sec then deccelarates from there on down and is at -48mm/s^2 at 4 sec plugged in. is this correct? for displacement would it be S 0-1= a/2 t^2= 12-6t/2 X t^2 = 12-6(1)/x 1^2= 3mm s 2-4 = a/2 t^2 = 12-6t/2 x t^2 = 12-6(4)/2 X (4)^2= 96mm s = 3mm + 96mm = 102mm for total displacement? or is that completely wrong?
 Feb 15th 2011, 06:15 AM #4 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 I believe your integral is wrong. Also, as you were told, the maximum velocity occurs when a = 0. 0 = 12 - 6t t = 2 s Then, we come to the velocity function. v = \int 12 - 6t dt v = 12t - (6t^2)/2 dt v = 12t - 3t^2 From there, can you work out the maximum velocity? Displacement involves yet another integral, but this time: s = \int 12t - 3t^2 dt Find the value of s when t = 4. Post what you get! __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Feb 15th 2011, 06:01 PM #5 Junior Member   Join Date: Feb 2011 Posts: 3 i got 12mm max and 36mm for total disp
 Feb 16th 2011, 04:36 AM #6 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 The maximum velocity is good, but I got a different displacement... could you show your calculations? __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?

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