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

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