Crossing the River.

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Presentation transcript:

Crossing the River

Your boat can only travel 10 m/s (that is how fast it moves it still water).

You want to cross a river that flows 3 m/s West to East and is 100 m wide.

If you want to make it across in the shortest amount of time, should you: aim straight across (point your bow North)? aim upstream by 30o (30o West of North)? aim downstream by 30o (30o East of North)? wait until the next day when there will be no current?

Solve the time it takes to cross for each of the above. Solve the displacement for each of the above. Where do you aim in order to arrive directly North of your starting position?

Solutions 1) In determining the time it takes to cross, you must remember that because the distant to be covered is 100 m [N], you need to use the “North” component of your velocity to determine the time. Because the current is flowing East, it has no North component.

∆d = 100 m [N] VNorth = 10 m/s [N] Therefore, time = 10 s

∆d = 100 m [N] VNorth = 10 m/s (cos300) = 8.7 m/s [N] Therefore, time = 11.5 s

∆d = 100 m [N] VNorth = 10 m/s (cos300) = 8.7 m/s [N] Therefore, time = 11.5 s

2) To determine the displacement, you need to put the velocity vector of your boat tip=to-tail with the velocity vector of the current and get the resultant velocity. Use that with the time from question 1 to get the displacement.

vr = 10.4 m/s [17o E of N] Time = 10 s Therefore, ∆d = 104 m [17o E of N]

vr = 8.9 m/s [13o W of N] Time = 11.5 s Therefore, ∆d = 102.4 m [13o W of N]

vr = 11.8 m/s [43o E of N] Time = 11.5 s Therefore, ∆d = 135.7 m [43o E of N]

To have a resulting velocity straight North, you need to aim upstream so that you form a right triangle with the 10 m/s as the hypotenuse. Angle = [17.5o W of N]