After flying for 19 min in a wind blowing 49 kmh at an angle

After flying for 19 min in a wind blowing 49 km/h at an angle of 23° south of east, an airplane pilot is over a town that is 57 km due north of the starting point. What is the speed of the airplane relative to the air?

Solution

We will consider N as the +y and E as the +x direction then we can represent the widn velocity in vector form as

V_w = 49*Cos(23) i^ - 49*Sin(23) j^

        = 45.10 i^ - 19.15 j^

Speed of the plane = 57/19 *60 = 180 km/hr, it is due north , writing it in vector form

Velocity of the airplane

V_a = 180j^

Velocity of the airplane is result of the wind speed and the speed of the plane relative to the wind, let us call the relative speed as V_r then we will have

V_a   = V_w + V_r

V_r = V_a - V_w = 180j^ - (45.10 i^ - 19.15 j^) = -45.10i^ + 199.15 j^

Magnitude of the speed = sqrt(45.10² + 199.15² ) = 204.19 km/hr

speed of the plane relative to the wind = 204.19 km/hr