Question

A light plane attains an airspeed of 500 km/h. The pilot sets out for a destination 780 km due north but discovers that the plane must be headed 19.0° east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude and (b) direction of the wind velocity? Give the direction as an angle relative to due west, where north of west is a positive angle, and south of west is a negative angle.

Answer 1

The component of the air speed in the north direction = 500*cos(19) = 472.76 km/hr

but since it takes 2 hours then the effective north velocity is 780/2 =390 km/hr

so the north component of the wind is 390 - 472.76 = -82.76 km/hr

The east component of the plane is 500*sin(19) = 162.78km/hr

so the west component of the wind is 162.78 km/hr

a)Therefore the wind speed is sqrt(82.76^2 + 162.78^2) = 182.6 km/hr

and

b)the direction is S of W

and angle = arctan(-82.76 /162.78) = -26.95 degrees