resolve the two vectors into x and y components
add the respective components
find the resultant vector
An airplane is traveling at 650 mph on a heading of 255°. The wind is blowing from a bearing of 320° at 28 mph. What is the ground speed and the actual bearing of the airplane?
P.S. I prefer the law of cosine/sine method
3 answers
All angles are measured CW from +Y-axis.
Vp = 650[255o] + 28[320o].
X = 650*sin255 + 28*sin320 =
Y = 650*Cos255 + 28*Cos320 =
Vp = sqrt(X^2+Y^2). = Velocity of the plane(gnd. speed).
Direction: Tan A = X/Y.
t
Vp = 650[255o] + 28[320o].
X = 650*sin255 + 28*sin320 =
Y = 650*Cos255 + 28*Cos320 =
Vp = sqrt(X^2+Y^2). = Velocity of the plane(gnd. speed).
Direction: Tan A = X/Y.
t
Vp = 650[255o]
Correction: Since the wind is blowing FROM 320o, the heading is 140o.
Vp = 650mi/h[255o] + 28mi/h[140o].
X = 650*sin255 + 28*sin140 = -609.9 mi/h.
Y = 650*Cos255 + 28*Cos140 = -189.7 mi/h.
Vp = -609.9 - 189.7i = 639mi/h[72.7o] W. of S. = 639mi/h[252.7o] CW.
Tan A = X/Y.
=
Correction: Since the wind is blowing FROM 320o, the heading is 140o.
Vp = 650mi/h[255o] + 28mi/h[140o].
X = 650*sin255 + 28*sin140 = -609.9 mi/h.
Y = 650*Cos255 + 28*Cos140 = -189.7 mi/h.
Vp = -609.9 - 189.7i = 639mi/h[72.7o] W. of S. = 639mi/h[252.7o] CW.
Tan A = X/Y.
=
3 answers