# Vector Question

1. Sep 13, 2007

### Eathrock

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

A carrier drives from her office 22.0 km northerly direction to town. Drives in direction of 60.0 degrees south of east for 47.0 km to another town. Whats her displacement from office?

2. Relevant equations

3. The attempt at a solution

I drew a diagram of her route but I do not know what to do next.

2. Sep 13, 2007

### learningphysics

What type of shape do you get?

3. Sep 13, 2007

### Eathrock

A Traingle

4. Sep 13, 2007

### learningphysics

cool. so do you know which side it is whose length you need? do you know the cosine rule?

5. Sep 13, 2007

### Eathrock

Hypotunese is what I need, right?
Cosine rule is to determine the sides/anlges of a non right angle.
Could I use Pythagorean theorem?

Last edited: Sep 13, 2007
6. Sep 13, 2007

### learningphysics

In this case you don't have a right triangle, so no hypoteneuse... so you can't use pythagorean theorem. But you can use the cosine rule to get the length of the third side of that triangle...

Then you need the direction also... so you need the angle of that third side (measured from north, south, east or west... whichever is most convenient).

7. Sep 13, 2007

### Eathrock

Okay. Let me try to work it out.

8. Sep 13, 2007

### Eathrock

a^2=b^2+c^2-2bccosA

9. Sep 13, 2007

### Eathrock

40.731

10. Sep 13, 2007

### learningphysics

careful... what angle do you have in the triangle... opposite the side you're calculating?

11. Sep 13, 2007

### Eathrock

I did a^2 = 22^2 + 47^2 - 2(22)(47)cos60 to get 40.731 and then I did
40.731^2 = 47^2 + 22^2 - 2(47)(22)cosB to get me 60

12. Sep 13, 2007

### learningphysics

Yes, but in my diagram, the 60degrees, isn't the angle in the triangle... I get 90-60=30 degrees...

13. Sep 13, 2007

### Eathrock

How do you get 90 degrees?

14. Sep 13, 2007

### learningphysics

90 degrees is the angle between the path (22.0 km north) and the east axis... she drives 60 degrees south of east... that means her path is 60 degrees below the east axis... or in other words 30 degrees east of south...

For example... 10 degrees south of east, is 80 degrees east of south...

15. Sep 13, 2007

### Eathrock

Oh okay. So I should replace 60 with 30 in the formula I showed you earlier.

16. Sep 13, 2007

### learningphysics

17. Sep 13, 2007

### Eathrock

Ok. I see. Make so much more sense now.
Is the side 30.530 and is that the answer?

18. Sep 13, 2007

### learningphysics

I'm getting 30.034km... you should also get the direction.

19. Sep 13, 2007

### Eathrock

Yes. You are right. 30.034 km.

How would you find the direction?

20. Sep 13, 2007

### learningphysics

You can find the angle of the displacement from any of the 4 axes (north south east west). I've drawn in one here as x:

http://www.imagevimage.com/gallery.php?entry=images/triangle2.gif

So when you find x... the direction would be x degrees west of south... hope that makes sense...

To find x... you'd use the sine law...

I've drawn x in two places... because those two angles are equal (since the lines are parallel)... hope this makes sense...