Fundamentals of Physics Extended (10th Edition)

$x_{(4)}=64m$
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 $x_{(4)}=64m$