Post of a function is s = t^3 - 6t^2 + 9t + 1
When is the particle at rest?
Can someone do a step by step for me...I have a test soon so I kinda just need to know.
The particle is at rest when the velocity is 0.
$\displaystyle v = \frac{ds}{dt} = 3t^2 - 12t + 9$
$\displaystyle 0 = 3t^2 - 12t + 9$
$\displaystyle 0 = 3t^2 - 9t - 3t + 9$
$\displaystyle 0 = 3t(t - 3) - 3(t - 3)$
$\displaystyle 0 = (t - 3)(3t - 3)$
The particle is at rest at $\displaystyle t = \{1, 3\}$.