\section*{Problem 1 - 031041}

A cyclist is riding over a bridge at a constant speed. When he has covered \(\frac{3}{8}\) of the distance, he meets another cyclist coming towards him at the same speed.

At what speed were both cycling if a car traveling at \(80 \frac{\mathrm{km}}{\mathrm{h}}\) on the same road met one at the beginning and the other at the end of the bridge?
}

Designations: 1. Cyclist: \(x\), 2. Cyclist: \(y\), Length of the bridge \(b\), Speed of the cyclists: \(v_{x}=v_{y}\) and of the car: \(v_{A}\), Distance: \(s\), Time: \(t\)

At time \(t_{0}\), \(y\) starts and at time \(t_{1}\), \(x\) starts. At time \(t_{2}\), \(x\) and \(y\) meet, i.e., \(s_{x, t_{2}}+s_{y, t_{2}}=b\). Since \(s_{x, t_{2}}=\frac{3}{8} b\), it follows that \(s_{y, t_{2}}=\frac{5}{8} b\).

Finally, at time \(t_{3}\), \(y\) reaches the destination, so \(s_{y, t_{3}}=b\) and thus \(s_{x, t_{3}}=\frac{6}{8} b\). \(x\) reaches the other side of the bridge at time \(t_{4}\): \(s_{x, t_{4}}=b\), i.e., \(s_{x, t_{4}-t_{3}}=\frac{1}{4} b\).

Case 1: The car drives in the same direction as \(x\), thus starts at \(t_{3}\) and arrives at \(t_{4}\), covering a distance of \(s_{A, t_{4}-t_{3}}=b\), i.e., it holds that

\[
v_{A}=\frac{s_{A, t_{4}-t_{3}}}{t_{4}-t_{3}}=\frac{b}{t_{4}-t_{3}} \quad \text { and } \quad v_{x}=\frac{s_{x, t_{4}-t_{3}}}{t_{4}-t_{3}}=\frac{\frac{b}{4}}{t_{4}-t_{3}}
\]

Thus, \(v_{x}=\frac{v_{A}}{4}=20 \frac{\mathrm{km}}{h}\).

Case 2: The car drives in the opposite direction to \(x\), thus starts at \(t_{0}\) and arrives at \(t_{1}\) and again covers a distance of \(s_{A, t_{1}-t_{0}}=b\), i.e., it holds that

\[
v_{A}=\frac{s_{A, t_{1}-t_{0}}}{t_{1}-t_{0}}=\frac{b}{t_{1}-t_{0}} \quad \text { and } \quad v_{y}=\frac{s_{y, t_{1}-t_{0}}}{t_{1}-t_{0}}=\frac{\frac{b}{4}}{t_{1}-t_{0}}
\]

Thus, it also follows that \(v_{y}=v_{x}=\frac{v_{A}}{4}=20 \frac{\mathrm{km}}{h}\).