The angle of the airplane’s path is = [tex](\frac{250\sqrt{2}+55}{\sqrt{2}} )[/tex]î +[tex](\frac{250\sqrt{6}-55 }{2})[/tex]ĵ
A vector is defined as a quantity which has two independent properties magnitude and direction.
î & ĵ denote two unit vectors along East & North directions respectively.
V1 denotes velocity vector direction of airplan of 30 degrees South of East at 500 mph
V2 denotes velocity vector wind the direction of 45 degrees North of East at 55 mph
V denotes the actual velocity vector of the given plane.
Hence from above data we get following descriptions,
V1 = 500.(sin 30°).î + 450.(cos 30°).ĵ
= 500.(1/2).î + 500(√3/2.ĵ
= 250î + 250√3ĵ
V2 = 55(sin 45°).î - 55(cos 45°).ĵ
= 55[1/√2].î - 55[1/√2]ĵ
V = V1 + V2
= 250î + 250√3ĵ +55[1/√2].î - 55[1/√2]ĵ
= 250î +55[1/√2].î + 250√3ĵ - 55[1/√2]ĵ
= [tex](\frac{250\sqrt{2}+55}{\sqrt{2}} )[/tex]î +[tex](\frac{250\sqrt{6}-55 }{2})[/tex]ĵ
Hence the angle of the airplane’s path is = [tex](\frac{250\sqrt{2}+55}{\sqrt{2}} )[/tex]î +[tex](\frac{250\sqrt{6}-55 }{2})[/tex]ĵ
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