Answer to Question #96041 in Mechanics | Relativity for sarah

Question

A jet airliner moving initially at 689 mph
(with respect to the ground) to the east moves
into a region where the wind is blowing at
424 mph in a direction 59◦
north of east.
What is the new speed of the aircraft with
respect to the ground?
Answer in units of mph.

Answer 1

The angle between two vectors is

$$\theta=59\degree$$

The new velocity of the aircraft with respect to the ground

$$\bold{V=v_a+v_w}$$

The new speed of the aircraft with respect to the ground:

$$V=\sqrt{v_a^2+v_w^2+2v_av_w\cos{59\degree}}$$

$$V=\sqrt{689^2+424^2+2(689)(424)\cos{59\degree}}=978\ mph$$