#### Answer

$$ \frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89} $$

#### Work Step by Step

Since
\begin{aligned}
\frac{d s}{d t} &=\sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} \\
&=\sqrt{9 t^{4}-24 t^{2}+16+4 t^{2}} \\
&=\sqrt{9 t^{4}-20 t^{2}+16}
\end{aligned}
Now, we will find the minimal speed,
$$f(t) =\sqrt{9 t^{4}-20 t^{2}+16} $$
Then $$ f'(t) =\frac{36 t^{3}-40 t }{2\sqrt{9 t^{4}-20 t^{2}+16} }$$
Hence $ f'(t) =0 $ at $t=0 $ and $t= \sqrt{\frac{10}{9}}$ then
$$\frac{d s}{d t}\bigg|_{t=0}= 4,\ \ \frac{d s}{d t}\bigg|_{t=0}=\sqrt{4.89}$$
Hence
$$ \frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89} $$