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A sailboat is sailing due east at 8.20 mph. The wind appears to blow from the southwest at 10.9 mph.
What is the true wind speed?

What is the true wind speed if the wind appears to blow from the northeast at 10.9 mph?

2007-03-15 03:51:51 · 2 answers · asked by RelientKayers 4 in Science & Mathematics Physics

2 answers

I would look it up in google

2007-03-15 03:54:22 · answer #1 · answered by amnon w 1 · 0 2

These are vector problems. You have to add corresponding x and y components of boat and apparent velocity vectors to get true velocities. There's another wrinkle: "from the northeast" = "toward the southwest", and vice-versa. In the answers, a wind direction of 0 deg is from the North, and angle increases clockwise, as is traditional in reporting wind speed (see ref.). However, the vector math is most easily done in the conventional algebraic system where velocities are in the "to" direction, 0 deg is along the +x or East axis and angle increases counterclockwise. When you calculate a resultant direction theta in the conventional algebraic system, you can convert to the "traditional wind" system by subtracting theta from 270 degrees. If the result is negative, add 360 degrees.
Here are the solutions:

Apparent from the southwest:
Vwind = Vsailboat + Vapparentwind
Vsailboat = [8.2,0]
Vapparentwind = [-10.9*sqrt(.5), -10.9*sqrt(.5)]
Vwind = [8.2-10.9*sqrt(.5), -10.9*sqrt(.5)] = [.493, -7.71]
Wind speed is 7.723 mph. Wind direction is (from) 244.15 deg or roughly out of the West-Southwest.

Apparent from the northeast:
Vwind = Vsailboat + Vapparentwind
Vsailboat = [8.2,0]
Vapparentwind = [10.9*sqrt(.5), 10.9*sqrt(.5)]
Vwind = [8.2+10.9*sqrt(.5), 10.9*sqrt(.5)] = [15.91, 7.71]
Wind speed is 17.68 mph. Wind direction is (from) 356.34 deg or roughly out of the North.

2007-03-15 11:55:16 · answer #2 · answered by kirchwey 7 · 0 0

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