\section*{Problem 4 - V01104}

A \(90 \mathrm{~m}\) long D-train is traveling at a speed of \(72 \mathrm{~km} \cdot h^{-1}\). It is \(150 \mathrm{~m}\) away from the railway crossing when a cyclist, who is \(100 \mathrm{~m}\) away from the railway crossing, notices it, moving towards the crossing at a speed of \(6 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

After how many seconds will the cyclist have the smallest distance from the end of the train?
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Assuming the railway crossing is perpendicular to the railway track, we consider a coordinate system where the railway crossing is at the origin and the train moves along the x-axis, while the cyclist moves along the y-axis.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1658.jpg?height=342&width=1037&top_left_y=1231&top_left_x=504)

At time \(t=0 \mathrm{~s}\), the cyclist is \(100 \mathrm{~m}\) away from the crossing, and the end of the train (train distance + train length) is \(240 \mathrm{~m}\). The position of the end of the train is described by the coordinates \((240 + 20 \cdot t)\), where \(t\) is measured in seconds. The speed of the train is \(72 \mathrm{~km} / \mathrm{h} = 20 \mathrm{~m} / \mathrm{s}\).

The position of the cyclist is analogously \((0, 6 \cdot t - 100)\).

The distance between the cyclist and the end of the train is

\[
d = \sqrt{(240 + 20 \cdot t)^2 + (6 \cdot t - 100)^2}
\]

The distance \(d\) is minimized when the radicand is minimized. The radicand is described by

\[
f(t) = 4(109 t^2 - 2700 t + 16900)
\]

The first derivative is \(f'(t) = 8(109 t - 1350)\) with a zero at \(t_1 = \frac{1350}{109} \approx 12.39\). Since the radicand is a quadratic function with a positive coefficient for the quadratic term, \(t_1\) is the time at which the distance between the end of the train and the cyclist is minimized. After 12.39 seconds, the distance is minimal.