Answer
$v_{avg}=18\frac{m}{s}$
Work Step by Step
As $x_{(t)}=12t^2-2t^3$:
For $t=0s$:
$x_1=x_{(0)}=12(0)^2-2(0)^3=0m$
For $t=3s$:
$x_2=x_{(3)}=12(3)^2-2(3)^3=54m$
Thus, we obtain:
$v_{avg}=\frac{x_2-x_1}{t_2-t_1}=\frac{54-0}{3-0}$
$v_{avg}=18\frac{m}{s}$
