Question

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.

Answer 1

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

$cos(\alpha) = 80/320 = 0.25$

$\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0$ relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

$320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h$

Answer 2

Answer:

a) Ф=14°, north of west

b) 310 km/h

Explanation:

we have,

$v_{p/G}$, velocity of plane relative to the ground(west)

$v_{p/A}$,velocity of plane relative to the air(320 km/h)

$v_{A/G}$,velocity of air relative to the ground(80 km/h, due south).

                 $v_{p/G}$=$v_{p/A}$+$v_{A/G}$.........(1)

a)     sin(Ф)=$\frac{v_{A/G}}{v_{p/A}}$

                 =$\frac{80km/h}{320km/h}$              

Ф=14°, north of west

b)    using Pythagorean theorem

$v_{p/G}=\sqrt{v^2_{p/A}+v^2_{A/G}}$

$v_{p/G}$=310 km/h

note:

diagram is attached