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

`301.29\ \text{mph}`

`119.5^(\circ)`

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=√(v_p^2+v_w^2+2v_pv_w\cos\theta)\\\Rightarrow v_r=\sqrt{320^2+35^2+2* 320* 35\cos125^(\circ)}\\\Rightarrow v_r=301.29\ \text{mph}`

Direction is given by

`\phi=\tan^(-1)(v_p\sin\theta)/(v_w+v_p\cos\theta)\\\Rightarrow \phi=\tan^(-1)(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)`.