\section*{Problem 3 - 011013}

A train travels over a \(171 \mathrm{~m}\) long bridge at a low speed in \(27 \mathrm{~s}\) (calculated from the moment the locomotive enters the bridge until the last car leaves the bridge). The train passes a pedestrian walking towards it at a speed of \(1 \mathrm{~m} / \mathrm{s}\) in \(9 \mathrm{~s}\).

a) What is the speed of the train (in \(\mathrm{km} / \mathrm{h}\))?

b) How long is the train?
}

The train passes the pedestrian at a speed \(v\) in \(9 \mathrm{~s}\), during which the pedestrian walks \(9 \mathrm{~m}\) towards the train. The train thus covers a distance of \(s-9 \mathrm{~m}\), where \(s\) is the length of the train. Therefore,

\[
v=\frac{s-9 \mathrm{~m}}{9 \mathrm{~s}}
\]

Since the train covers a distance of \(171 \mathrm{~m}+s\) in 27 s, we also have

\[
v=\frac{171 \mathrm{~m}+s}{27 \mathrm{~s}}
\]

By equating the two equations and solving for \(s\), we obtain \(s=99 \mathrm{~m}\). Substituting this value, we get \(v=10 \frac{\mathrm{m}}{\mathrm{s}}=36 \frac{\mathrm{km}}{\mathrm{h}}\).