3. (18 points) As shown in Figure 6, in trapezoid $ABCD$, $AD \parallel BC$, $AB \perp BC$, $AD = 8 \, \text{cm}$, $BC = 16 \, \text{cm}$, $AB = 6 \, \text{cm}$. Moving points $M$ and $N$ start from points $B$ and $C$ simultaneously, moving along $BC$ and $CD$ respectively,
B
The speeds of points $M$ and $N$ are $2 \, \text{cm/s}$ and $1 \, \text{cm/s}$ respectively.
(1) After how many seconds have points $M$ and $N$ moved such that $MN \parallel BD$?
(2) While point $M$ is moving on side $BC$, let the time point $M$ has moved be $t \, \text{s}$. Is there a moment $t \, \text{s}$ such that the area of $\triangle AMN$ is minimized? If so, find the value of $t$; if not, explain why.
3. (1) Suppose points $M$ and $N$ have moved for $x \, \text{s}$, such that $M N \parallel B C$. Then
$$
B M=2 x(\text{cm}), C M=16-2 x(\text{cm}), C N=x(\text{cm}).
$$

From the given information, $B D=10(\text{cm})$.
Draw $D H \perp B C$ at point $H$, then $B H=A D=8$.
Thus, $C H=8(\text{cm}), C D=B D=10(\text{cm})$.
Therefore, $D N=10-x(\text{cm})$.
When $\frac{C M}{B M}=\frac{C N}{D N}$, $M N \parallel B D$.
From $\frac{16-2 x}{2 x}=\frac{x}{10-x}$, we get $x=\frac{40}{9}(\text{s})$.
(2) It exists.
$$
\begin{array}{l}
=\frac{1}{2}(8+16) \times 6-\frac{1}{2} \times \\
{\left[6 \times 2 t+(16-2 t) \times \frac{3}{5} t+8\left(6-\frac{3}{5} t\right)\right]} \\
=72-\left(6 t+\frac{24}{5} t-\frac{3}{5} t^{2}+24-\frac{12}{5} t\right) \\
=\frac{3}{5}\left(t^{2}-14 t\right)+48 \\
=\frac{3}{5}(t-7)^{2}+\frac{93}{5}(0<t<8) \text{. } \\
\end{array}
$$

When $t=7 \, \text{s}$, $\dot{S}_{\triangle \mathrm{Mev}}$ is minimized, and the minimum value is $\frac{93}{5}(\text{cm}^{2})$. (Provided by Feng Kaike)