The Classic Airplane Problem

AxeluteZero

1. The problem statement, all variables and given/known data

A small plane departs from point A heading for an airport 490 km due north at point B. The airspeed of the plane is 210 km/h and there is a steady wind of 50 km/h blowing directly toward the southeast.
(a) Determine the proper heading for the plane.
° west of north

(b) How long will the flight take?
h

2. Relevant equations

Law of Sines/Cosines?

v_f = v_i + a*t


3. The attempt at a solution

So I drew a picture.

You need a velocity vector. This vector must be the sum of the velocity vectors of the wind and the plane, which should also be drawn on the sketch. So I drew a simple sketch showing the plane's northward trek, an angle [tex]\Theta[/tex], another vector showing the wind and the resulting vector. So:

The first vector points due North (up), from point A.
The wind vector points toward point A, at an angle [tex]\Theta[/tex].
The resultant vector points from point B to the non-pointed end of the wind vector.

...so where in the world does the trig come in? I just don't know where to start plugging in numbers, since this isn't a right triangle it isn't a simple sin/cos problem with the Pythag. Theorem, right?

rl.bhat

In this problem, AB is the resultant velocity (Vr) of wind velocity (Vw)and proper velocity of the plane (Vp)
To make the problem easy, consider wind direction as x-axis.
Angle between wind direction and the resultant is ( 90 + 45) degrees.
Let θ be the angle between Vp and Vw. Then if you draw the vectors, you can see that,
Vr*sin(135) = Vp*sinθ .....(1)
Vr*cos(135) = Vp*cosθ - Vw .......(2)
Now solve for Vr and θ.

AxeluteZero

Ok I solved for Vr and for theta, but I can't use either of them since there's still the other variable in the equation. IE theta is still in the Vr equation (and I don't know it) and I have Vr in the theta equation. And plugging either equal to each other makes it unsolvable. I end up with a cos (-) plus sin (-).

Vr = Vr

[tex]\frac{210 * sin \Theta}{sin (135)}[/tex] = [tex]\frac{210 (cos\Theta) - 50}{cos 135}[/tex]

Help?

rl.bhat

Divide eq. 1 by 2 . You get
(Vp*sinθ) = Vp*cosθ - Vw
Square both the sides. You get
Vp^2*sin^2(θ) =( Vp*cosθ - Vw)^2
Vp^2*[1- cos^2(θ)] = ( Vp*cosθ - Vw)^2
Now simplify and solve the quadratic to find cosθ. Then find Vr.