1. Two cars start from the same location at the same time, driving in the same direction at the same speed. Each car can carry a maximum of 24 barrels of gasoline. They cannot use any other fuel during the journey. Each barrel of gasoline can make a car travel 60 kilometers. Both cars must return to the starting point, but they do not have to return at the same time. They can lend fuel to each other. To make one of the cars travel as far away from the starting point as possible, at what distance from the starting point should the other car turn back? How many kilometers in total has the car that travels the farthest traveled?
If so, A gives B $(24-2 x)$ barrels of gasoline. B continues to move forward, carrying $(24-2 x)+(24-x)=48-3 x$ barrels of gasoline. According to the problem, we should have
$$
48-3 x \leqslant 24 \text {. That is, } x \geqslant 8 \text {. }
$$

After A and B separate, the distance B continues to travel is
$$
\begin{array}{l}
S(x)=-\frac{(24-2 x)+(24-2 x)}{2} \cdot 60 \\
=30(48-4 x) \text {. } \\
\end{array}
$$

Therefore, when $x=8$, $S(x)$ is maximized $=S(8)=480$ (km).
Thus, the total distance B travels is $2[60 x+480]_{2}=8$ $=1920$ (km).