\section*{Problem 4 - 081024}

A car is driving at a constant speed on a straight road from A to B. A second car is driving on the same road at a constant speed from B to A.

Both cars start this journey at the same time in A and B, respectively. They meet each other at a certain point on the road.

After the meeting, the first car still has exactly \(2 \mathrm{~h}\) to drive to B, and the second car still has exactly \(\frac{9}{8}\) h to drive to A. The distance between A and B (measured on the road) is 210 km.

Determine the speeds of the cars!
}

The speed values (in \(\frac{\mathrm{km}}{\mathrm{h}}\)) of the two motor vehicles \(K_{1}, K_{2}\) are \(v_{1}, v_{2}\); the time values (in \(\mathrm{h}\)) in which they travel the distance \(A B\) are \(t_{1}, t_{2}\).

Then (1) \(210=v_{1} t_{1}=v_{2} t_{2}\).

Furthermore, both \(\left(t_{1}-2\right) \mathrm{h}\) and \(\left(t 2-\frac{9}{8}\right) \mathrm{h}\) are the times from the start of the journey until the meeting, so (2) \(t_{1}=t_{2}+\frac{7}{8}\) must hold.

Finally, the distance from the meeting point \(T\) to \(B\) is \(\left(v_{1} \cdot 2\right) \mathrm{km}\) and from \(T\) to \(A\) is \(\left(v_{2} \cdot \frac{9}{8}\right) \mathrm{km}\), from which (3) \(v 1 \cdot 2+v_{2} \cdot \frac{9}{8}=210\) follows.

The solution of the system of equations (1), (2), (3) is obtained with

\[
\begin{array}{r}
210 \cdot t_{2} \cdot 2+210 \cdot\left(t_{2}+\frac{7}{8}\right) \cdot \frac{9}{8}=210 \cdot\left(t_{2}+\frac{7}{8}\right) \cdot t_{2} \\
t_{2}^{2}-\frac{9}{4} t_{2}-\frac{63}{64}=0 \\
t_{2_{1 ; 2}}=\frac{9}{8} \pm \frac{1}{8} \sqrt{81+63}=\frac{9}{8} \pm \frac{12}{8}
\end{array}
\]

Of these, only \(t_{2}=\frac{21}{8}\) is usable, since \(t_{2}>0\). From (2) and (1), it follows further that \(t_{1}=\frac{7}{2}\) and \(v_{1}=60 \frac{\mathrm{km}}{\mathrm{h}}\) as well as \(v_{2}=80 \frac{\mathrm{km}}{\mathrm{h}}\)
\footnotetext{
Solution adopted from \([5]\)
}

\subsection*{7.10.3 Third Round 1968, Class 10}