\section*{Task 2 - 011112}

A steamboat travels from A to B in 3 hours and, with the same engine power, from B to A in \(4 \frac{1}{2}\) hours. How long does a vessel propelled only by the current take for the journey from A to B?
}

We can assume the distance-time law of uniform motion \(s=vt\) in both directions of the river. For the trip downstream, this means \(v=v_{D}+v_{S}\), and for the trip upstream, \(v=v_{D}-v_{S}\), where \(v_{D}\) is the boat's own speed and \(v_{S}\) is the river's current speed. Therefore,

\[
\begin{array}{r}
s=\left(v_{D}+v_{S}\right) \cdot(3 \text{ h})=\left(v_{D}-v_{S}\right) \cdot(4.5 \text{ h}) \\
v_{D}=\frac{4.5 \text{ h}+3 \text{ h}}{4.5 \text{ h}-3 \text{ h}} v_{S}=5 v_{S}
\end{array}
\]

and thus \(s=\left(v_{D}+v_{S}\right) \cdot(3 \text{ h})=6 v_{S} \cdot(3 \text{ h})\). For a boat that is only carried by the current, \(s=v_{S} t\); with the above equation, we have

\[
s=6 v_{S} \cdot(3 \text{ h})=v_{S} t
\]

From this, the travel time for a vessel driven only by the current is \(t=18 \text{ h}\).