Finding the Correct Compass Heading The pilot of an aircraft wishes to head directly east but is faced with a wind speed of $40mi/hr$from the northwest. If the pilot maintains an airspeed of $250mi/hr$, what compass heading should be maintained to head directly east? What is the actual speed of the aircraft?

Consider that the pilot of an aircraft wishes to head directly east, but is faced with a wind speed of $40mi/hr$from the northwest. If the pilot maintains an airspeed of $250mi/hr$, it asks to determine compass heading should be to maintain to head directly east and the actual speed of the aircrafi Now we set up a coordinate system in which north (N) is along the positive y- axis as shown in above figure.

NORTH

$40mi/hr$

$v_g$

$30°\theta 250mi/hr$

WEST EAST
SOUTH

Further the velocity of the wind$v_w$has magnitude 40 and direction$N30°W$(northwest).So the angle between$v_w$and horizontal componentiis$90°+30°=120°$.
Thus a compass heading of 120 degree should be maintained to head directly east.

Again we express $v_w$in terms of i and j as

$$\begin{gathered} v_a=250(\cos 0°i+\sin 0°j) \\ =500(1\times i+0\times j) \\ =500i \end{gathered}$$

Velocity of the 737 relative to the ground vg is always given as,

$$\begin{gathered} v_R=v_a+v_w \\ =500i-20i+20\sqrt{3}j \\ =480i+20\sqrt{3}j \end{gathered}$$

$$\begin{gathered} \left|\right|v_R\left|\right|=\sqrt{{\left(480\right)}^2+{\left(20\sqrt{3}\right)}^2} \\ =\sqrt{230400+1200} \\ =\sqrt{231600} \\ \approx 481.2milesperhour \end{gathered}$$