Answer to Question #94634 in Mechanics | Relativity for Ali hamza

Question

A wind is blowing at 50 km h−1
from a direction 45° west of north. The pilot of an airplane wishes to fly on
a route due north from an airport. The airspeed of the airplane is 250 km h−1
.
(a)In what direction must the pilot point the nose of the airplane?
(b) What will be the airplane’s speed relative to the ground?

Answer 1

Resulting aircraft speed:

$\vec{v} = \vec{u} + \vec{w},$

where $\vec{u}$ is the own aircraft airspeed, $\vec{w}$ is the wind speed.

Since wind is blowing in 45 degrees angle,

$w_y = w\cos45\degree = w_x = w\sin45\degree.$

Projecting the vectors on axes in absolute values:

• in north(+)/south(-) axis: $v_y = u_y - w_y = u\cos\alpha - w\cos45\degree;$
• in east(+)/west(-) axis: $0 = u_x-w_x = u\sin\alpha - w\sin45\degree .$

(here $v_x=0$ because we know the aircraft in result is heading due north).

From the 2nd equation,

$\sin\alpha = \frac{w}{u}\sin45\degree; \\ \alpha = \arcsin(\frac{w}{u}\sin45\degree) = \arcsin(\frac{50\frac{\text{km}}{\text{h}}}{250\frac{\text{km}}{\text{h}}}\frac{\sqrt2}{2}) \approx 8.13\degree.$

The above angle is the direction of pointing the nose of the aircraft, west of north.

From the 1st equation,

$v = v_y = u\sqrt{1-\sin^2\alpha} - w\cos45\degree = u\sqrt{1-(\frac{w}{u}\sin45\degree)^2} - w\cos45\degree = \\ = 250\frac{\text{km}}{\text{h}}\sqrt{1-(\frac{50\frac{\text{km}}{\text{h}}}{250\frac{\text{km}}{\text{h}}}\frac{\sqrt2}{2})^2} - 50\frac{\text{km}}{\text{h}}\frac{\sqrt2}{2} \approx 212.13 \frac{\text{km}}{\text{h}}.$