One. (20 points) Two cars, A and B, start from the same point $A$ at the same time, driving in the same direction in a straight line. Each car can carry a maximum of $240 \mathrm{~L}$ of gasoline, and they cannot refuel during the journey. Each liter of gasoline allows a car to travel $12 \mathrm{~km}$. Both cars must return to the starting point along the same route, but they can borrow gasoline from each other. Design a scheme to allow one of the cars to travel as far away from the starting point $A$ as possible, and calculate the total distance this car has traveled.
Let the car A travel $x \mathrm{~km}$ and car B travel $y \mathrm{~km}$, as far away from the starting point $A$ as possible. Then
$$
\left\{\begin{array}{l}
x+y \leqslant 240 \times 12 \times 2, \\
x-y \leqslant 240 \times 12 .
\end{array}\right.
$$

and $x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y) \leqslant 4320$, which means car A has traveled a total of $4320 \mathrm{~km}$.

The specific plan is: after both cars have traveled $720 \mathrm{~km}$, car B gives car A $60 \mathrm{~L}$ of gasoline and waits at this location. Car A continues to travel another $1440 \mathrm{~km}$ before returning. When car A meets car B, it borrows another $60 \mathrm{~L}$ of gasoline, and then both cars return to the starting point $A$.