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
resolve the two vectors into x and y components
add the respective components
find the resultant vector
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]
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.
=
To solve this problem using the law of cosines and sines, we can break it down into two parts: finding the resultant velocity of the airplane in terms of its ground speed and heading, and then finding the actual bearing of the airplane.
First, let's find the resultant velocity of the airplane. This can be done by finding the horizontal and vertical components of the airplane's velocity.
The horizontal component of the airplane's velocity is given by:
Vx = ground speed * cos(heading)
The vertical component of the airplane's velocity is given by:
Vy = ground speed * sin(heading)
We can now calculate the horizontal and vertical components of the wind velocity as well, given the bearing of the wind and its speed:
Wind horizontal component:
Wx = wind speed * cos(wind bearing)
Wind vertical component:
Wy = wind speed * sin(wind bearing)
Now, the resultant horizontal component of the airplane's velocity is given by:
Rx = Vx + Wx
And the resultant vertical component of the airplane's velocity is given by:
Ry = Vy + Wy
To find the ground speed, we can use the Pythagorean theorem:
ground speed = sqrt(Rx^2 + Ry^2)
To find the heading or actual bearing of the airplane, we can use the inverse tangent function:
actual bearing = arctan(Ry / Rx)
Substituting the given values into the formulas, we have:
Vx = ground speed * cos(255°)
Vy = ground speed * sin(255°)
Wx = 28 mph * cos(320°)
Wy = 28 mph * sin(320°)
Now we can plug these values into the formulas and solve for ground speed:
Rx = Vx + Wx
Ry = Vy + Wy
ground speed = sqrt(Rx^2 + Ry^2)
Once the ground speed is found, we can then calculate the actual bearing using:
actual bearing = arctan(Ry / Rx)
Using these formulas and performing the necessary calculations, we can determine the ground speed and actual bearing of the airplane.