An airplane leaves the airport when there is an Easterly wind (i.e a wind blowing from direction East) of speed 43 km/h. If the plane starts out directly above Taupo with an airspeed v=140 km/h, pointing in a direction θ = 26° East of North, how far is it from Taupo after 63 min of flying at constant altitude?
(Ans: 134 km)
The Attempt at a Solution
I found the x and y components of the velocity:
Vx=140 cos 26 = 125.83
Vy=140 sin 26 = 61.37
Since the wind is blowing from the East at a speed of 43 km/h:
Vx=125.83 - 43 = 82.83 km/h
I could now use the formula vx=d/t, but I don't think the problem is asking for just the displacement in the x direction. I'm very confused. Can anyone show me what I need to do?