## pathosdebater 3 years ago 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?

1. ivanmlerner

What do you mean by Ax, I didn't get it

2. pathosdebater

Ax is the magnitude of the original vector of the plane.

3. ivanmlerner

Yes, I read it, but what does this vector mean?

4. pathosdebater

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.

5. pathosdebater

It is asking for the length of the unknown side

6. ivanmlerner

|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)

7. hubertH

|dw:1351980839247:dw| use cosine rule to get the unknown side

8. hubertH

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)