Answer
$ -12 m/s^2 $
Work Step by Step
The position is given by $ x = 12t^2 - 2t^3 $
Since the velocity equation is the derivative of the position equation we need to find
$\frac{d}{dt} (12t^2 - 2t^3) $
Since $\frac{d}{dt} t^n = n*t^(n-1)$ then
$\frac{d}{dt} (12t^2 - 2t^3) = 12(2)t - 2(3)t^2 = 24t - 6t^2 $
Therefore $v = 24t - 6t^2$
The acceleration is given by the derivative of the velocity equation
This is $ \frac{d}{dt} (24t - 6t^2) = 24 - 6(2)t = 24 - 12t $
Therefore, to find acceleration at t =3, we plug in t = 3 into the equation
This gives us $ 24 -12(3) = 24-36 = -12 $ Therefore the acceleration is $-12m/s^2$
