A plane flies on a heading of S60 degrees East at a constant speed of 550 km/h.
If the velocity of the wind is 50 km/h on a bearing of S40 degrees West, what is the velocity of the plane with respect to the ground?
3 answers
the resultant velocity is just the sum of the two vectors. To get that, convert each vector to x- and y- coordinates, add them, then convert back to polar form.
That is if you know if the terminology means the wind is coming from the southwest or heading to the southwest. Bearing does not mean heading or direction of travel. I have no idea which it means. For me a SW wind means from the SW headed NE
v = 550(cos330,sin330) + 50(cos230,sin230)
= (476.3139..., -275) + (-32.1393..., -38.30222...)
= (444.17459..., -313.022..)
|v| = √(444.17459...^2 + (-313.022..)^2) = appr 543.55
or by cosine Law after making your sketch and using the diagonal of the parallelogram.
v^2 = 550^2 + 50^2 - 2(550)(50)cos80°
= 543.55..
= (476.3139..., -275) + (-32.1393..., -38.30222...)
= (444.17459..., -313.022..)
|v| = √(444.17459...^2 + (-313.022..)^2) = appr 543.55
or by cosine Law after making your sketch and using the diagonal of the parallelogram.
v^2 = 550^2 + 50^2 - 2(550)(50)cos80°
= 543.55..
2 answers