Question

If a particle moves along a straight line according to `s(t)=t^4-4t^3+6x^2-20`, how do you find the maximum & minimum velocity on `0<=t<=3`?

Answer 1

Please see the explanation below

Explanation:

The position is given by

`s(t)=t^4-4t^3+6t^2-20`

The velocity is the derivative of the position

`v(t)=s'(t)=4t^3-12t^2+12t`

To find the max and min velocity, differentiate the velocity

`a(t)=v'(t)=12t^2-24t+12`

The critical points are when

`v'(t)=0`

`12t^2-24t+12=12(t^2-2t+1)=0`

`(t-1)^2=0`

The solutions are

`{(t=1),(4t^3-12t^2+12t=0):}`

At `t=1`, there is an inflection point.

Let's make a variation chart

`color(white)(aaaa)` `t` `color(white)(aaaaaa)` `-oo` `color(white)(aaaaaaa)` `1` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `v'(t)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `v(t)` `color(white)(aaaaaaaa)` `nn` `color(white)(aaaa)` `4` `color(white)(aaaa)` `uu`

So, when `t in [1,3]`

The minimum is `v(0)=0`

The maximum is `v(3)=36`

graph{4x^3-12x^2+12x [-13.37, 15.11, -4.05, 10.2]}