Question

Using the derivative, when is the particle at rest? (Velocity = 0)

Answer 1

We have the following equation

$s(t)=t^3-6t^2+1$

if we derive it we get

$s^{\prime}(t)=3t^2-12t$

Now, let's set this equation to zero and solve for t

$3t^2-12t=0$

using the quadratic formula

$\begin{gathered} ax^2+bx+c=0 \\ x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}$$t_{1,\: 2}=\frac{-\left(-12\right)\pm\sqrt{\left(-12\right)^2-4\cdot\:3\cdot\:0}}{2\cdot\:3}$$t_1=\frac{-\left(-12\right)+12}{2\cdot\:3},\: t_2=\frac{-\left(-12\right)-12}{2\cdot\:3}$

the solutions to the quadratic equation are:

$t=4,\: t=0$

Thus, the particle is at rest at t=0 and t=4