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
Stacey Warren - Expert brainly.com
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test that and the endpoints
Plug in 0, 8/6, and 4 into 3t^2-8t-3 and see where you get the smallest value.
0 should be the min acceleration.. i think
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
@ExperimentX he's asking for the MINIMUM of ACCELERATION
you gave him the MAXIMUM of VELOCITY
Oo mistake .. for minimun you get at t = 3, where v'' has negative value
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?
So when u're differentiating acceleration, you're actually taking the 2nd derivative of velocity.
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