\(s\left(t\right)=-t^3+6t^2\)
=>\(v\left(t\right)=-3\cdot t^2+6\cdot2t=-3t^2+12t\)
\(\Leftrightarrow a\left(t\right)=v'\left(t\right)=-3\cdot2t+12=-6t+12\)
\(v\left(t\right)=-3t^2+12t\)
\(=-3t^2+12t-12+12\)
\(=-3\left(t-2\right)^2+12< =12\forall t\)
Dấu '=' xảy ra khi t=2
\(a\left(2\right)=-6\cdot2+12=0\)