## Fundamentals of Physics Extended (10th Edition)

$v_{avg}=18\frac{m}{s}$
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: $v_{avg}=\frac{x_2-x_1}{t_2-t_1}=\frac{54-0}{3-0}$ $v_{avg}=18\frac{m}{s}$