Question

A plane is flying at a heading of 125° at a speed of 320 mph. The wind is blowing from due North at a speed of 35 mph. Find the ground speed and the course of the plane. (Note: The course must be between 0° and 360°, where due North corresponds to a course of 0°, due East to a course of 90°, due South to a course of 180°, and due West to a course of 270°.)

Answer 1

Answer:

$301.29\ \text{mph}$

$119.5^{\circ}$

Step-by-step explanation:

$v_p$ = Velocity of plane = 320 mph

$v_w$ = Velocity of wind = 35 mph

$\theta$ = Angle between wind and plane directions = $125^{\circ}$

From triangle law we have resultant

$v_r=\sqrt{v_p^2+v_w^2+2v_pv_w\cos\theta}\\\Rightarrow v_r=\sqrt{320^2+35^2+2\times 320\times 35\cos125^{\circ}}\\\Rightarrow v_r=301.29\ \text{mph}$

Direction is given by

$\phi=\tan^{-1}\dfrac{v_p\sin\theta}{v_w+v_p\cos\theta}\\\Rightarrow \phi=\tan^{-1}\dfrac{320\sin125^{\circ}}{35+320\cos125^{\circ}}\\\Rightarrow \phi=-60.46^{\circ}=180-60.46=119.5^{\circ}$

The magnitude of the plane is $301.29\ \text{mph}$ moving at an angle of $119.5^{\circ}$.