Given that, $S=t^5-40 t^3+30 t^2+80 t-250$
$$
\begin{aligned}
& V=\frac{d S}{d t}=5 t^4-120 t^2+60 t+80 \\
& a=\frac{d V}{d t}=\frac{d^2 S}{d t^2}=20 t^3-240 t+60
\end{aligned}
$$

Let
$$
\begin{aligned}
a & =f(t)=20 t^3-240 t+60 \\
f^{\prime}(t) & =60 t^2-240 \\
f^{\prime}(t) & =0 \Rightarrow 60 t^2-240=0 \Rightarrow t^2=4 \\
t & = \pm 2 \Rightarrow t=2, t=-2 \\
f^{\prime \prime}(t) & =120 t
\end{aligned}
$$
at $t=2, \quad f^{\prime \prime}(t)>0$
So, at $t=2, f(t)$ is minimum
So, minimum acceleration is
$$
\begin{aligned}
a_{\min } & =f(2)=20(2)^3-240 \cdot 2+60 \\
& =160-480+60=-260 .
\end{aligned}
$$