Answer
$t = \sqrt (4/3)$ and t = $-\sqrt(4/3)$
Work Step by Step
The position is given by $ x = 20t - 5t^3 $
To find the velocity, we need to take the derivative of the position equation or
$ \frac{d}{dt} (20t - 5t^3) $
The position is given by $ x = 20t - 5t^3 $
Since the velocity equation is the derivative of the position equation we need to find
$\frac{d}{dt} (20t - 5t^3) $
Since $\frac{d}{dt} t^n = n*t^(n-1)$ then
$ v = 20 - 5(3)t^2 = 20 - 15t^2 $
To find when the velocity is 0, we plug in 0 for v and solve for t
This gives us
$ 0 = 20 - 15t^2 $ so then $ 20 = 15t^2 $ or $ (4/3) = t^2 $
Taking the square root of both sides we find that $ t = \sqrt (4/3)$ and $-\sqrt(4/3)$
