Posted by **Willie** on Monday, September 21, 2009 at 7:39pm.

Two boats, A and B, travel with a velocity of 4.90 m/s across a river of width 72.0 m. The river flows with a velocity of 2.50 m/s. Boat A travels the shortest distance and boat B travels in the shortest time. If both start at the same time, how much time will they take to cross the river?

- Physics -
**MathMate**, Monday, September 21, 2009 at 9:15pm
The shortest time is to orient the boat at 90° to the shore. However, the boat will be carried downstream on arriving the other side.

The time required will be the width of the river divided by the velocity of the boat in still water.

The shortest distance is also the only way by which the boat arrives directly opposite to the starting point.

The boat must be oriented towards upstream in such a way that the component parallel to the river cancels out the velocity of the river, or

4.9cos(θ)=2.5

where θ is the angle with the river.

Solving, &theta=59.3226° with the river.

The component of the boat's velocity perpendicular to the river is thus

4.9sin(θ) m/s

The time required is therefore

Width of river / (4.9sin(θ)) s.

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