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)=(-b\pm√(b^2-4ac))/(2a) \end{gathered}`
`t_(1,\: 2)=(-\left(-12\right)\pm√(\left(-12\right)^2-4\cdot\:3\cdot\:0))/(2\cdot\:3)`
`t_1=(-\left(-12\right)+12)/(2\cdot\:3),\: t_2=(-\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