as the fellow above said, the particle does make a U-turn at the vertex, and we can find that out by getting the derivative of s(t) and zeroing it out to get the point of the horizontal tangent.
Bearing in mind that ds/dt is really v(t) or the velocity equation, so when we zero out the ds/dt, is the same as saying the velocity went to 0, since it got zeroed out, and then the particle goes over the vertex and changes direction.
[tex]\bf s(t)=2t^3-21t^2+60t+3\implies \boxed{\cfrac{ds}{dt}=6t^2-42t+60}\leftarrow v(t)
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0=6t^2-42t+60\implies 0=t^2-7t+10\implies 0=(t-5)(t-2)
\\\\\\
t=
\begin{cases}
5\\
2
\end{cases}\impliedby \textit{it makes the first turn at the \underline{2 second}}
\\\\\\
\textit{now let's find the acceleration equation, a(t)}
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\boxed{\cfrac{d^2s}{dt^2}=12t-42}\leftarrow a(t)\\\\
-------------------------------\\\\
s(2)=55\qquad \qquad \qquad \qquad a(2)=-18[/tex]
the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.
