Four. (This question is worth 20 points) A point moves at a speed of 2 meters/second on the $x$-axis, and at a speed of 1 meter/second in other parts of the plane. Try to find the boundary line of the region that the point can reach within 1 second from the origin.


By symmetry, we only need to consider the boundary points $(a, b)$ in the first quadrant. Clearly, $0 \leqslant a \leqslant 2$.
$b$ is the maximum ordinate of the point that can be reached on the line $x=a$ within 1 second.

Let the last point on the $x$-axis of the most time-saving route from $(0,0)$ to $(a, b)$ be $(x, 0)(0 \leqslant x \leqslant a)$. Then we have
$$
\frac{x}{2}+\sqrt{b^{2}+(a-x)^{2}}=1 \text {, }
$$

which implies $b^{2}=-\frac{3}{4} x^{2}+(2 a-1) x+1-a^{2}$.
Next, we find the maximum value of $b^{2}$.
Let $f(x)=-\frac{3}{4} x^{2}+(2 a-1) x+1-a^{2}$, with its axis of symmetry at $x=\frac{2}{3}(2 a-1)$, and $0 \leqslant x \leqslant a$.

When $\frac{2}{3}(2 a-1) \leqslant 0$, i.e., $a \leqslant \frac{1}{2}$, $f(0)$ is the maximum, i.e., the maximum value of $b^{2}$ is $1-a^{2}$.
When $0<\frac{2}{3}(2 a-1) \leqslant a$, i.e., $\frac{1}{2}<a \leqslant 2$, $f\left(\frac{2}{3}(2 a-1)\right)$ is the maximum, i.e., the maximum value of $b^{2}$ is $\frac{1}{3}(a-2)^{2}$.
Thus, the boundary of the region is
$$
|b|=\left\{\begin{array}{ll}
\sqrt{1-a^{2}}, & |a| \leqslant \frac{1}{2} \text { when, } \\
\frac{1}{\sqrt{3}}|2-| a||, \frac{1}{2} \leqslant|a| \leqslant 2 \text { when. }
\end{array}\right.
$$