The compass in a light aircraft shows that it is flying due north. The air speed indicator gives a reading of 150km/h. If there is a 50km/h wind blowing from west to east, what is the ground speed of the aircraft?

2. Originally Posted by bobby77
The compass in a light aircraft shows that it is flying due north. The air speed indicator gives a reading of 150km/h. If there is a 50km/h wind blowing from west to east, what is the ground speed of the aircraft?
The plane is moving 150km/h forward, it is also moving 50km/h right. If you draw out the vectors, as shown in the diagram below, then you'll see we want to calculate the hypotenuse of the triangle.

use pythagorean theorum...
$c=\sqrt{150^2+50^2}$

$c=\sqrt{22500+2500}$

$c=\sqrt{25000}$

$c=50\sqrt{10}\approx158.114$

3. It is always good to specify direction.
Direction will be $tan^{-1}(\frac{1}{3})$east of north.

Malay

4. ## A question by itself

I'm just learning vectors mind you...

would the velocity of the ground speed of the aircraft be $50\bold{i}+150\bold{j}+0\bold{k}$ or $50\bold{i}+0\bold{j}+150\bold{k}$?

5. Originally Posted by Quick
I'm just learning vectors mind you...

would the velocity of the ground speed of the aircraft be $50\bold{i}+150\bold{j}+0\bold{k}$ or $50\bold{i}+0\bold{j}+150\bold{k}$?
It depends on your coordinate system. You draw any three mutually perpendicular lines, and your coordinate system is ready(anyone of the three could be x,y and z). But for international convention, a coordinate system is universally accepted(right handed coordinate system).
Draw a horizontal line and a line perpendicular to it(it will be obviously vertical).Label horizontal as x , vertical as y. Then z will be upwards out of the plane of the paper.
In the question
velocity of plane (without air)=0i+150j+0k
velocity of air=50i+0j+0k
resultant velocity=50i+150j+0k
(Actually there is a method which could tell you that the cordinate system used by you is acc. to international convention or not(based on cross product of vectors))