II. (25 points) A straight road $l$ runs through a grassland, and there is a health station $A$ by the road. A residential point $B$ is located $30 \mathrm{~km}$ away from the road, and the distance between $A$ and $B$ is $90 \mathrm{~km}$. One day, a driver is tasked with delivering emergency medicine from the health station to the residential point. The car's maximum speed on the road is 60 $\mathrm{km} / \mathrm{h}$, and its maximum speed on the grassland is $30 \mathrm{~km} / \mathrm{h}$. What route should the driver take to minimize the travel time? What is the shortest travel time?
As shown in Figure 8, let the route on the road be $A D$, and the route on the grass be $D B$, with the distance of $A D$ being $x$. Given the conditions $A B=90$, $B C=30$, and $B C \perp A C$, we have
$$
A C=\sqrt{A B^{2}-B C^{2}}=60 \sqrt{2} \text {. }
$$

Then, $C D=A C-A D=60 \sqrt{2}-x$,
$$
B D=\sqrt{C D^{2}+B C^{2}}=\sqrt{30^{2}+(60 \sqrt{2}-x)^{2}} \text {. }
$$

Let the total travel time be $y$, then
$$
y=\frac{x}{60}+\frac{\sqrt{30^{2}+(60 \sqrt{2}-x)^{2}}}{30} \text {. }
$$

Rearranging into a quadratic equation in terms of $x$, we get
$$
3 x^{2}+(120 y-480 \sqrt{2}) x-3600 y^{2}+3600 \times 9=0 \text {. }
$$

Since $x$ must exist, $\Delta \geqslant 0$, i.e.,
$$
[120(y-4 \sqrt{2})]^{2}-4 \times 3 \times 3600\left(9-y^{2}\right) \geqslant 0 \text {. }
$$

Simplifying, we get $4 y^{2}-8 \sqrt{2} y+5 \geqslant 0$.
Solving, we find $y \geqslant \sqrt{2}+\frac{\sqrt{3}}{2}$ or $y \leqslant \sqrt{2}-\frac{\sqrt{3}}{2}$ (discard the latter).
Therefore, the minimum value of $y$ is $\sqrt{2}+\frac{\sqrt{3}}{2}$. At this point,
$$
A D=x=60 \sqrt{2}-10 \sqrt{3} \text {. }
$$