[tex]s(t)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t[/tex]
To find when the particle is at rest, equal the velocity function (derivate of the position function) to 0 and solve t:
Velocity function:
[tex]\begin{gathered} v(t)=\frac{d}{\text{ d t}}(\frac{2}{3}t^3-\frac{9}{2}t^2-18t) \\ \\ v(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18(1) \\ \\ v(t)=2t^2-9t-18 \end{gathered}[/tex]
Equal to 0 the equation above:
[tex]2t^2-9t-18=0[/tex]
Solve t:
[tex]\begin{gathered} \text{Factor:} \\ 2t^2+3t-12t-18=0 \\ t(2t+3)-6(2t+3)=0 \\ (t-6)(2t+3)=0 \\ \\ t-6=0 \\ t=6 \\ \\ 2t+3=0 \\ 2t=-3 \\ t=-\frac{3}{2} \end{gathered}[/tex]
Solutions for t are: t=6 and t=-3/2.
As the time cannot be a negaive amount. The particle is at rest at a time of 6
