Please help... A plane heads west (180.0 degrees) at a speed of 400.0 mph. The wind's velocity is southeast (315.0 degrees) at 30. mph.
What is Ax, the magnitude of the original vector of the plane?
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What do you mean by Ax, I didn't get it
Ax is the magnitude of the original vector of the plane.
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Well I'm guessing that the equation forms a triangle. One side is west (180.0 degrees) at a speed of 400.0 mph and another is southeast (315.0 degrees) at 30. mph. I think Ax is the final vector of the triangle.
It is asking for the length of the unknown side
|dw:1351909145239:dw|Using vectors, we see that the velocity vector has value 400 in the direction of x, and the wind velocity vector has value -cos(45°)*30 in the direction of x.
The y component of the wind velocity vector is sin(45°)*30
Now, we have that the vector you want is: (400, 0)-(-cos(45°)*30, sin(45°)*30)=(400+sqrt(2)*15, sqrt(2)*15) and the magnitude of the vector is: sqrt((400+sqrt(2)*15)^2+2*15^2)
|dw:1351980839247:dw| use cosine rule to get the unknown side
the answer shoul be 379,38mph
Don't forget when using direction headings, North is 0 degrees amd South is 180 degrees, and west is 270 degrees (not 180 degrees)
Please help... A plane heads west (270.0 degrees) at a speed of 400.0 mph. The wind's velocity is southeast (135.0 degrees) at 30. mph.
Would be describing the situation and the wind direction is in the direction that it is going. Not coming from as is the normal reporting of wind.