2. As shown in Figure 2, person A at point $A$ on the shore notices person B in distress at point $B$ in the water. Point $B$ is 30 meters away from the shore at $C$, and $\angle BAC=15^{\circ}$. Person A runs on the shore at a speed that is $\sqrt{2}$ times their swimming speed in the water. Given that A's swimming speed in the water is 3 meters/second, the time it takes for A to reach B from point $A$ is $t$ seconds. Then the minimum value of $t$ is $\qquad$
2. $15 \sqrt{2}+5 \sqrt{6}$

As shown in Figure 3, construct
$\angle C A D=45^{\circ}$, and let point $P$ be where person 甲 enters the water from the shore. Draw $P H \perp A D$ at point $H$, at this time,
$$
A P=\sqrt{2} P H .
$$

According to the problem, the time it takes for 甲 to travel from point $A$ to $B$ is the time spent swimming the distance $P H + P B$ in the water.

Draw $B E \perp A D$ at point $E$. By the shortest distance from a point to a line, we know $P H + P B \geqslant B E$, with equality holding if and only if point $E$ coincides with $H$.
In the right triangle $\triangle A B C$,
$$
A B=\frac{B C}{\sin \angle C A B}=\frac{30}{\sin 15^{\circ}} \text {. }
$$

In the right triangle $\triangle A E B$,
$$
\begin{array}{l}
B E=A B \sin (\angle D A C+\angle C A B) \\
=45 \sqrt{2}+15 \sqrt{6} .
\end{array}
$$

Therefore, $t_{\min }=15 \sqrt{2}+5 \sqrt{6}$.