A circular racing track is $1 \mathrm{~km}$ long. Two motorcycles start simultaneously from a given point $A$ on the track, in opposite directions. One travels at a constant speed, while the other accelerates uniformly. They first meet at point $B$ on the track, and meet again at point $A$. How far did the first motorcycle travel from the start to point $B$?
Let the speed of the first motorist be \( v \) (expressed in \(\mathrm{km} / \mathrm{h}\)), and the acceleration of the second motorist be \( a \) (for consistency, the unit will be \(\left.\mathrm{km} / \mathrm{h}^{2}\right)\). Since both cover one kilometer in the same time \( t \), we have \(\frac{a}{2} t^{2} = v t = 1\), which means \( t = \frac{1}{v} \) and \( a = \frac{2 v}{t} = 2 v^{2} \).

Let the time elapsed until the meeting at point \( B \) be \( T \). We know that in time \( T \), the total distance covered by the two motorists is one kilometer, so

\[
\frac{a}{2} T^{2} + v T = 1
\]

Substituting the value of \( a \),

\[
v^{2} T^{2} + v T - 1 = 0
\]

The \( v T \) in the equation represents the distance covered by the first motorist until the meeting at point \( B \) (in kilometers). The equation is quadratic in \( v T \), and since the distance can only be positive,

\[
v T = \frac{-1 + \sqrt{5}}{2}
\]

Based on the work of Gyurkó L. Gergely (Fazekas M. Főv. Gyak. Gimn., II. o.t.)