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Old Feb 10th 2011, 12:34 AM   #1
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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.
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Old Feb 10th 2011, 04:32 AM   #2
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Originally Posted by cramjm11 View Post
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.)

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Old Feb 15th 2011, 01:33 AM   #3
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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?
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Old Feb 15th 2011, 06:15 AM   #4
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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!
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Old Feb 15th 2011, 06:01 PM   #5
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i got 12mm max
and 36mm for total disp
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Old Feb 16th 2011, 04:36 AM   #6
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The maximum velocity is good, but I got a different displacement... could you show your calculations?
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