Solution:

We claim that Aerith can escape, even when Bob plays optimally and always runs towards Aerith's current location.
Let a dash (d) be a unit of distance and a tick ($t$) be a unit of time such that the radius of the lake is $9$ dashes and such that one mph is equal to one dash per tick $(\mathrm{d} / \mathrm{t})$. Let $\Omega$ be the perimeter of the lake, and let $\omega$ be a concentric circle with radius $1.98$ dashes.$^{1}$

b)

c)

We outline a four-step strategy for Aerith to escape the lake:

a) First, Aerith swims to the edge of $\omega$.

b) Then, Aerith begins swimming along $\omega$. Her angular velocity is then
$$\left(2 \frac{\mathrm{d}}{\mathrm{t}}\right) /(1.98 \mathrm{~d})>1 \mathrm{t}^{-1}$$
However, Bob's angular velocity is exactly $\left(9 \frac{d}{t}\right) /(9 \mathrm{~d})=1 \mathrm{t}^{-1}$, which is slightly less, so Aerith outpaces Bob. Thus after some amount of time, Aerith will eventually be at the furthest point on $\omega$ from Bob. At this point in time, let Aerith's current position be $A$ and let Bob's current position be $B$.

c) Aerith now swims outward on line $AB$ until she reaches the point $A'$ that is $2 \mathrm{~d}$ away from the center of the lake. This amount of time this takes is
$$(2 \mathrm{~d}-1.98 \mathrm{~d}) /\left(2 \frac{\mathrm{d}}{\mathrm{t}}\right)=0.01 \mathrm{t}$$
and in this time Bob travels $0.01t \cdot 9 \frac{\mathrm{d}}{\mathrm{t}}=0.09 \mathrm{~d}$.

d) Finally, Aerith swims away from Bob along the straight line through $A'$ perpendicular to line $AB$. Aerith's angular velocity along this journey is now less than $1 \mathrm{t}^{-1}$ because, as we saw earlier, she has to be within a circle of radius $2$ in order to outpace Bob. Therefore, she remains ahead of Bob, so Bob will continue in his direction when chasing her. It takes her $\left(\sqrt{(9 \mathrm{~d})^{2}-(2 \mathrm{~d})^{2}}\right) /\left(2 \frac{\mathrm{d}}{\mathrm{t}}\right)=\frac{\sqrt{77}}{2} \mathrm{t}$ to reach the edge, and in this time Bob can travel $\frac{\sqrt{77}}{2} \mathrm{t} \cdot 9 \frac{\mathrm{d}}{\mathrm{t}}=\frac{9 \sqrt{77}}{2} \mathrm{~d}$.

The total distance Bob can travel along the circle in the time it takes Aerith to reach the edge is thus $0.09 \mathrm{~d}+\frac{9 \sqrt{77}}{2} \mathrm{~d}<40 \mathrm{~d}$. However, in order to arrive on time to tag Aerith, he would need to travel $9\left(\frac{3 \pi}{2}-\arcsin \left(\frac{2}{9}\right)\right)>40 \mathrm{~d}$, so Aerith will indeed escape before Bob catches up.