Question

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

Answer 1

differentiate the acceleration

Answer 2

i think u meant v(t)=t^3-4t^2-3t+2

Answer 3

a=v'=3t^2-8t-3
a'=6t-8=0
6t=8
t=8/6

Answer 4

test that and the endpoints

Answer 5

Plug in 0, 8/6, and 4 into 3t^2-8t-3 and see where you get the smallest value.

Answer 6

0 should be the min acceleration.. i think

Answer 7

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

Answer 8

@ExperimentX he's asking for the MINIMUM of ACCELERATION

Answer 9

you gave him the MAXIMUM of VELOCITY

Answer 10

Oo mistake .. for minimun you get at t = 3, where v'' has negative value

Answer 11

haha .. .plus it was outside boundary

Answer 12

experiment stop confusing ppl lol

Answer 13

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?

Answer 14

So when u're differentiating acceleration, you're actually taking the 2nd derivative of velocity.

Answer 15

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

Answer 16

no, because u have to test endpoints too

Answer 17

plug them all in into acceleration equation.