13.350. Two motorists met halfway between cities $A$ and $B$. At the meeting, it turned out that the first one from $A$ had left earlier than the second one from $B$ by as many hours as half the time (also in hours) that would have passed until their meeting if they had left from the same points at the same time, on the same road, at the same speeds, constant throughout the journey. How many times faster was the second motorist driving compared to the first?
Solution.

Let $V_{1}, V_{2}$ be the speeds of the cars from cities $A$ and $B$ respectively, and $x$ be the distance between cities $A$ and $B$. Then the driver from $A$ left

$\frac{x}{2\left(V_{1}+V_{2}\right)}$ (hours) earlier than the driver from $B$ and traveled $\frac{x V_{1}}{2\left(V_{1}+V_{2}\right)}$ (km).

By the condition, $\frac{\frac{x}{2}-\frac{x V_{1}}{2\left(V_{1}+V_{2}\right)}}{V_{1}}=\frac{\frac{x}{2}}{V_{2}}$, from which $\frac{V_{2}}{V_{1}}=\frac{1+\sqrt{5}}{2}$.

Answer: in $\frac{1+\sqrt{5}}{2}$ times.