A pilot wants to fly north. The plane has an air speed of 350 km/h. There is a 50 km/h wind blowing to the west

Question

a. What direction does the plane need to head?

b. What is the plane’s velocity relative to the ground?

Answer 1

a.) In order to travel North we must convert some of our speed (350 km/hr) to the east in order to cancel the 50 km/hr wind speed to the west.

So if we think of this in terms of vectors, we know the x-component needs to be Vx = 50 km/hr East. The magnitude of our vector is the total speed of V = 350 km/hr. So to determine the direction we can use:

Cosθ = Vx/V

θ = Cos^-1(Vx/V) = Cos^-1(50 km/hr / 350 km/hr ) = 81.79° North of East.

b.) The ground speed would be the speed we appear to be traveling at in the North direction only (Vy) (since our heading accounted for the wind speed). So we will use:

Sinθ = Vy/V

VSinθ = Vy = 350 km/hr * Sin(81.79°) = 346.41 km/hr.