# 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#?

##### 1 Answer

Please see the explanation below

#### Explanation:

The position is given by

The velocity is the derivative of the position

To find the max and min velocity, differentiate the velocity

The critical points are when

The solutions are

At

Let's make a variation chart

So, when

The minimum is

The maximum is

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