# Obscure vector problem

1. Feb 6, 2009

### jemerlia

1. The problem statement, all variables and given/known data

To a cyclist X travelling at 8 km h-1 due east the wind appears to
blow from the south. Another cyclist Y travels at 16 km h-1 at
N30°W relative to X. The direction of the wind as experienced by Y
is from the west. Calculate the speed and direction of the wind.

2. Relevant equations

Vectors - relative motion

3. The attempt at a solution

The problem appears slightly ambiguous in that I would assume that N30W is an absolute direction. A vector diagram might be drawn:
.............................WY
.......................^--------->
....................../
....XG............./YX
^----------->/
|
|WX
|

where
WX is wind velocity relative to X
XG is X velocity relative to ground
YX is Y velocity relative to X
WY is wind velocity relative to Y

The vectors could be added WG = WY+YX+XG
but, of course, no magnitude is given for WY and WX - only the direction.

I can't see how to proceed from here - help gratefully received :)

2. Feb 6, 2009

### jemerlia

I wonder if taking Y's direction as an absolute value affects the problem - however I feel that if a compass bearing is quoted then it should be treated as a compass bearing - still puzzled by the word relative in "relative to X".

3. Feb 6, 2009

### LowlyPion

First of all you want to draw the cyclists as vectors traveling on their compass headings. X is easy.

X = 8 i + 0 j

However Biker Y is given relative to Biker X. This means that you need to account for Biker X's speed in determining Biker Y's true direction.

Y = - (16*Sin30° - 8) i + 16*Cos30° j

Then you need to figure what it means for Biker X to experience the wind as from the S and biker Y to experience it from the West.

4. Feb 7, 2009

### jemerlia

Thanks for the hint. What is true is that WG (wind speed relative to the ground) must be the same for both cyclists. Following your suggestion I drew separate vector diagrams for the two cyclists:

...............^.^
....WG.../......|
......./..........|WX WG = WX +XG
.../..............|
------------->
XG

WY
^----------->
..\.........../
.YG.\....../WG WG = WY+YG
........\../

It appears that the system might generate simultaneous equations allowing solution of j and i - but although I have given myself a 1h crash course in imaginary numbers I'm unsure of the next step and its mechanism - help gratefully received :)

5. Feb 7, 2009

### LowlyPion

I think that what you can conclude is that since each perceives only West or South and they are at right angles to each other, the Bicyclist 1 motion cancels the X component and perceives just the Y component (magnitude unknown) of the wind and Bicyclist 2's motion cancels the Y component of the wind and perceives only X directed wind. (I renamed them to avoid confusion.)

Hence taking the canceling components from their motions relative to the ground, the Wind is directed as = 8 i + 16*cos30 j

You can find |wind| through ordinary means.

6. Feb 7, 2009

### jemerlia

Please forgive my unfamiliarity with imaginary numbers. Do the i's and j's also represent the y and x co-ordinates? I am unsure how to solve the expression to find the co-ordinates.

7. Feb 7, 2009

### LowlyPion

So sorry. i,j's are simply vector notation for the vector components in the x,y directions respectively. If I could have written x-hat and y-hat more conveniently to emphasize the vector property I would have. It's just a short hand, and has nothing at all to do with imaginary numbers.

8. Feb 7, 2009

### jemerlia

Once again - thank you. I'm still puzzled as to how solve for two unknowns in the expression for the wind direction...

the Wind is directed as = 8 i + 16*cos30 j

9. Feb 7, 2009

### jemerlia

After further thought it was clear that your explanation was only a trivial step short of the answer. However, I would be grateful if you would please check that my reasoning is correct:

(a) Wind direction - the expression 8x + 16*cos30y (x,y instead of i,j) points to the co-ordinates (8, 16*cos 30). This yields an angle of 60degrees which corresponds with the given answer of blowing towards N30E.
(b) |wind| = SQRT(8^2 + (16*cos30)^2) = 16km/h - which again corresponds with the given answer.

10. Feb 7, 2009

### LowlyPion

While you do have 2 unknowns - namely the x-component and y component of wind - you can solve them independently by the fortuitous circumstance that the wind as perceived by 1 is only in one dimension at right angle to his motion (and at that he's cycling along an axis) and the wind perceived by 2 is at right angle to that.

Orthogonal is a wonderful thing.