Answer to Question #135433 in Mechanics | Relativity for Carol

Question

A jet airliner, moving initially at 300 Km/h to the east, suddenly enters a region where the wind is blowing at 100 Km/h toward the direction 30.00 north of east. What is the new speed and direction of the aircraft relative to the ground?

Answer 1

Solution

Given angle $\theta=72^0$

Speed of airliner

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

Where $v_a, v_w$ are speed of airliner and wind with respect to ground.

Therefore new speed of airliner ith respect to ground can be expressed as

$V=\sqrt{(300)^2+(100)^2+2\times (300)\times (100)cos30^°}$

$V=389.82Km/h$

Direction

$tan\theta=\frac{v_w}{v_a}=\frac{100}{300}=0.33$

$\theta= 18.26^°$