\section*{Task 2 - 011042}

On a river with a constant flow velocity \(v\), a motorboat travels downstream to a target that is a distance \(s\) away from the starting point, and then back again.

Another motorboat travels with the same own speed to a target that is also at a distance \(s\), but exactly perpendicular to the direction of the current, and then back again.

a) How much pure driving time do the two boats require?

b) What result is obtained for \(s=250 \mathrm{~m}, v=150 \frac{\mathrm{m}}{\mathrm{min}}\) and \(c=250 \frac{\mathrm{m}}{\mathrm{min}}\) ?
}

a) If the boat travels with the current, the speeds add up. If it travels against it, the difference is the effective speed. The travel time is the sum of the segment where the boat travels at \(c+v\) and the one at \(c-v\):

\[
t_{1}=\frac{s}{c+v}+\frac{s}{c-v}=\frac{2 v s}{c^{2}-v^{2}}
\]

If it moves perpendicular to the current, the speed of the river does not matter (it must of course be compensated for by the boat). Then \(t_{2}=\frac{2 s}{c}\).

b) After substituting the numerical values, we get: \(t_{1}=1 \frac{14}{30} \mathrm{~min}, \stackrel{\mathrm{im}}{\mathrm{c}}\) in the other case: \(t_{2}=2 \mathrm{~min}\).