What is the shortest distance you have to travel to reach where the other person was initially(12 ft apart)? You are 12ft away from a person walking towards you. He will keep walking straight, even in these COVID times. But you want to keep a minimum 6ft distance at all times.
Both are walking at the same speed at all times. And you can take any path. But he will move along the straight line joining the 2 points.
What is the shortest distance you have to travel to reach where the other person was initially (12 ft apart)?
 A: Here is a better solution than just walking along a half-circle of radius $6$:
Let $A$ and $I$ be the positions of the two people initially, and let the other person be at $I$, moving towards $A$ in a straight line. Then person $A$ can move along the path $AFJI$, as shown in the diagram:

Let $\angle FAB$ be a $60º$ angle. Therefore $\sin 60º = \frac{6}{AF}$, and $AF = \frac{6}{\sin 60º} = \frac{12 \sqrt 3}{3} = 4 \sqrt 3$. $J$ is a point which is defined by $CF + FJ = DB$ and $JI$ being tangent to the circle with centre $B$ and radius $6$. But $FJ$ and $JI$ are now hard to calculate. Therefore, let $J'$ be the point $(6,6)$, where the path now is longer than with $JI$. Then $FJ'$ is $3$ units long, and $IJ'$ is $6 \sqrt{2}$ units long, for a distance of at most $4 \sqrt{3} + 6 \sqrt{2} + 3 \approx 18.4$, less than $6 \pi \approx 18.8$. GeoGebra gives the total path length as approximately $17.79$.
For a demonstration that the distance between $A$ and $B$ is always greater than $6$ feet, check out this GeoGebra demonstration here.
A: The shortest distance is about 17.45.
Take the unit distance to be 6 feet, and let their speed also be unit.
Take $B$ to be at relative rest at $(0,0)$. In this picture, $A$ starts at the point $(-2,0)$ and ends at a point $(t,0)$ where $t$ is the final distance between $A$ and $B$, equal to the length actually travelled by $A$. From $B$'s viewpoint, the path taken should  be a straight line along a tangent to the unit circle, then an arc, and then another straight line to $C$. In the original true picture, $A$ moves along a line then along a curve that stays on the border of $B$'s moving unit circle and then moves in a straight line again (see diagram at bottom).

In the original picture, when $A$ moves at unit speed in direction $(\cos\theta,\sin\theta)$, then in the second picture, its velocity becomes the vector $$\begin{pmatrix}1+\cos\theta\\\sin\theta\end{pmatrix}=2\cos(\theta/2)\begin{pmatrix}\cos(\theta/2)\\\sin(\theta/2)\end{pmatrix}$$ Conversely, to go from the second picture to the original, the path's tangent vector of $(\cos\theta,\sin\theta)$ becomes $\frac{1}{2\cos\theta}(\cos2\theta,\sin2\theta)$.
There are four parts of the shortest path.

*

*The tangent at angle $30^\circ$ to the unit circle. Its length in the second picture is $\sqrt3$ but in the true picture it is $\frac{\sqrt3}{2\cos30^\circ}=1$ (in the direction of $2\times30^\circ=60^\circ$).


*The circular arc along the unit circle up to the vertical. Its true length is $$\int_0^{\pi/6}\frac{1}{2\cos\theta}d\theta=\tfrac{1}{2}[\ln(\sec\theta+\tan\theta)]_0^{\pi/6}=\tfrac{1}{4}\ln3$$


*The circular arc from the vertical by some angle $\theta$. Its true length is $$\tfrac{1}{2}\log(\sec\theta+\tan\theta)$$


*The straight line to $C$. Its true length is $\cot\theta/2\cos\theta=1/2\sin\theta$.
The final condition is that the total length of the path is $BC=t=1/\sin\theta$.
$$1+\tfrac{1}{4}\log3+\tfrac{1}{2}\log(\sec\theta+\tan\theta)+\frac{1}{2\sin\theta}=\frac{1}{\sin\theta}$$
This can be solved numerically: $\theta\approx0.351068$.
With this angle, the total length is $$\frac{1}{\sin\theta}=2.90782 \mathrm{units} = 17.4469 \mathrm{feet}$$
=================================================
Edit To clarify, when person $A$ comes at a distance of $1$ unit from $B$, they move along a curve that is not circular but is a translated version of the following curve: $$x'(t)^2+y'(t)^2=1,\qquad(x(t)-t)^2+y(t)^2=1,\quad (x(0),y(0))=(0,1)$$
The curve is related to the tractrix, except that it has unit speed all along. Only relative to a fixed $B$, does it appear circular.

A: Person A starts at point A and has to keep minimum of $6$ ft. distance from person B. Person B starts from point B and walks straight to point A. Both A and B have same speed.
Please see the diagram below. Path taken by person A is AD, DE and EB to get to point B.

$AC = 6, AD = 4 \sqrt3$.
He then walks distance $DE$ parallel to line $AB$ and point $E$ is such that $\angle CM'E = 60^0$ and $EM' = CD = 4 \sqrt3 - 6$.
So, $DE = CM' - (CD+EM').cos60^0 = 3 - (4 \sqrt3 - 6) = 9-4\sqrt3$.
Please note $AD + DE = 9$, so when person A is at point E, person B has already reached $C'$. So they have vertically crossed each other safely somewhere between point $D$ and $E$. Please also note that $C'M' = 6.$
This makes it safe for person A to now go straight from E to B.
$EE' = EM'.cos60^0 = 2 \sqrt3 - 3, EB' = 6 + EE' = 3 + 2 \sqrt3$
$BE = \sqrt{BB'^2+EB'^2} \approx 8.82$, and as obtained above, $AD + DE = 9$
So, total distance traveled by person A to reach point B keeping 6 ft. of min. distance from person B $\approx 17.82$.
