Answer
$x_{(4)}=64m$
Work Step by Step
When the velocity is zero, the maximum positive coordinate reached by particle occurs:
$v=24t-6t^2=0$
Factoring $t$ we find:
$t(24-6t)=0$
Thus, either $t=0 s$ or $24-6t=0$, which simplifies to $t=4s$
Now, we plug in the $t$ values into the position function
$x_{(t)}=12t^2-2t^3$
$x_{(0)}=12(0)^2-2(0)^3$
$x_{(0)}=0m$
$x_{(4)}=12(4)^2-2(4)^3$
$x_{(4)}=64m$
Thus, the maximum position of the particle is $x_{(4)}=64m$
