Answer to Question #136674 in Mechanics | Relativity for jjiud

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.

The task is shown in the following picture:

If we represent the motion of the aircraft as a vector $\vec a$ , the wind as a vector $\vec b$ and the new speed and direction of the aircraft as a vector $\vec c$ then:

$\vec c=\vec a+\vec b$ .

Find projections of the vectors on the axis x and y:

$x_{\vec a}=300;$ $y_{\vec a}=0.$

$x_{\vec b}=100\cdot cos(30\degree)=100\frac{\sqrt{3}}{2}=50\sqrt{3};$ $y_{\vec b}=100\cdot sin(30\degree)=100\cdot\frac{1}{2}=50.$

and $x_{\vec c}=x_{\vec a}+x_{\vec b}=300+50\sqrt{3}\approx386.60;$ $y_{\vec c}=y_{\vec a}+y_{\vec b}=0+50=50.$

Find the length of the vector $\vec c$ :

$|\vec c|=\sqrt{x_{\vec c}^2+y_{\vec c}^2}\approx\sqrt{386.60^2+50^2}\approx389.82$ Km/h.

Find the angle γ:

$γ=arcsin(\frac{y_{\vec c}}{|\vec c|})\approx arcsin(\frac{50}{389.82})\approx arcsin(0.1283)\approx7.37\degree$ .

Answer: the new speed and direction of the aircraft is 389.82 Km/h toward the direction 7.37° north of east.