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October 24, 2014

October 24, 2014

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

- Physics -
**MathMate**, Monday, September 21, 2009 at 9:15pmThe 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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