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?

May 6, 2018

Explanation:

The position is given by

$s \left(t\right) = {t}^{4} - 4 {t}^{3} + 6 {t}^{2} - 20$

The velocity is the derivative of the position

$v \left(t\right) = s ' \left(t\right) = 4 {t}^{3} - 12 {t}^{2} + 12 t$

To find the max and min velocity, differentiate the velocity

$a \left(t\right) = v ' \left(t\right) = 12 {t}^{2} - 24 t + 12$

The critical points are when

$v ' \left(t\right) = 0$

$12 {t}^{2} - 24 t + 12 = 12 \left({t}^{2} - 2 t + 1\right) = 0$

${\left(t - 1\right)}^{2} = 0$

The solutions are

$\left\{\begin{matrix}t = 1 \\ 4 {t}^{3} - 12 {t}^{2} + 12 t = 0\end{matrix}\right.$

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

Let's make a variation chart

$\textcolor{w h i t e}{a a a a}$$t$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$v ' \left(t\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$v \left(t\right)$$\textcolor{w h i t e}{a a a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$\cup$

So, when $t \in \left[1 , 3\right]$

The minimum is $v \left(0\right) = 0$

The maximum is $v \left(3\right) = 36$

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