## Task 2 - 010832

Peter has set up a "railway oval" on a board for his model railway (see the sketch; the circular arcs are semicircles).

Hans, whom he has invited, suddenly asks: "Do you think the train will go as fast as in real life?" Peter replies: "Certainly not, just imagine a real train next to it! Our train can hardly manage one kilometer in the 'Yes," says Peter, "that's true, but $1 \mathrm{~km}$ means something completely different for the layout

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0577.jpg?height=329&width=536&top_left_y=1160&top_left_x=1271)
to convert it." They think about it and then determine the following values:

| Time for one lap: | $11 \mathrm{~s}$ |
| :--- | :--- |
| Track gauge of the model railway: | $18.5 \mathrm{~mm}$ |
| Track gauge in reality: | $1435 \mathrm{~mm}$ |

a) How fast is the train actually going?

b) How fast would the train be from the perspective of the model railway?
a) To determine the actual speed of the train, one needs the length it covers on the rail oval. This consists of two straight sections and two semicircular arcs: $2 \cdot 700 \mathrm{~mm} + 2 \pi \cdot 400 \mathrm{~mm} = 3.91 \mathrm{~m}$.

This results in a speed of the model railway train of $35.6 \frac{\mathrm{cm}}{\mathrm{s}} = 1.28 \frac{\mathrm{km}}{\mathrm{h}}$.
b) A track gauge of $18.5 \mathrm{~mm}$ on the model railway corresponds to $1435 \mathrm{~mm}$ in reality. Therefore, the speed of the train in reality corresponds to

$$
\frac{35.6 \frac{\mathrm{cm}}{\mathrm{s}} \cdot 1435 \mathrm{~mm}}{18.5 \mathrm{~mm}} = 27.6 \frac{\mathrm{m}}{\mathrm{s}} \approx 100 \frac{\mathrm{km}}{\mathrm{h}}
$$