anonymous
  • anonymous
how do i find the minimum acceleration attained on the interval 0<=t<=4 by the particale whos velocity is given by v(t)=t^2-4t^2-3t+2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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bahrom7893
  • bahrom7893
differentiate the acceleration
bahrom7893
  • bahrom7893
i think u meant v(t)=t^3-4t^2-3t+2
bahrom7893
  • bahrom7893
a=v'=3t^2-8t-3 a'=6t-8=0 6t=8 t=8/6

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bahrom7893
  • bahrom7893
test that and the endpoints
bahrom7893
  • bahrom7893
Plug in 0, 8/6, and 4 into 3t^2-8t-3 and see where you get the smallest value.
bahrom7893
  • bahrom7893
0 should be the min acceleration.. i think
experimentX
  • experimentX
differentiate it, v' = 3*t^2 - 8*t - 3 find critical points, v' = 0 =? t = 3*t^2 - 8*t - 3 => t =3, -1/3 v '' = 6*t - 8 put the values of t at v'', where you get negative value you get maximun ie at t=1/3
bahrom7893
  • bahrom7893
@ExperimentX he's asking for the MINIMUM of ACCELERATION
bahrom7893
  • bahrom7893
you gave him the MAXIMUM of VELOCITY
experimentX
  • experimentX
Oo mistake .. for minimun you get at t = 3, where v'' has negative value
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experimentX
  • experimentX
haha .. .plus it was outside boundary
bahrom7893
  • bahrom7893
experiment stop confusing ppl lol
bahrom7893
  • bahrom7893
Okay here: You are given velocity, you need to find the min acceleration. Thus you will have to differentiate acceleration and find where its minimum occurs. But to get the formula of acceleration you must differentiate velocity. Got it?
bahrom7893
  • bahrom7893
So when u're differentiating acceleration, you're actually taking the 2nd derivative of velocity.
experimentX
  • experimentX
Oops sorry, i thought i should give the minimum value of velocity. okay that would be minimum value of v' right, that would be 8/6
bahrom7893
  • bahrom7893
no, because u have to test endpoints too
bahrom7893
  • bahrom7893
plug them all in into acceleration equation.

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