\section*{Problem 4 - 121034}

A locomotive driver noticed at the beginning of a \(20 \mathrm{~km}\) long section \(s\) that he was exactly 4 minutes late.

He then drove this section \(s\) at an average speed that was \(10 \mathrm{~km} / \mathrm{h}\) higher than what the timetable specified. At the end of section \(s\), the timetable was first met again.

What was the timetable-specified average speed for \(s\)?
}

For the distance, \(s=20 \mathrm{~km}\) applies. If \(v_{1}\) is the planned speed for the section \(s\) and \(t_{1}\) is the planned time, then

\[
v_{1}=\frac{20 \mathrm{~km}}{t_{1}}
\]

The locomotive driver travels at \(v_{2}=v_{1}+10 \frac{\mathrm{km}}{\mathrm{h}}\) and requires the time \(t_{2}\), which is \(4 \mathrm{~min}=\frac{1}{15} \mathrm{~h}\) shorter than \(t_{1}\). This results in

\[
v_{2}=v_{1}+10 \frac{\mathrm{km}}{\mathrm{h}}=\frac{20 \mathrm{~km}}{t_{1}-\frac{1}{15} \mathrm{~h}}=t_{2}
\]

Substituting (1) leads to the result \(t_{1}=\frac{2}{5} \mathrm{~h}=24 \mathrm{~min}\) and the planned speed \(v_{1}=\) \(50 \frac{\mathrm{km}}{\mathrm{h}}\). A check confirms the result.