1. Sep 30, 2007

### alandry06

1. The problem statement, all variables and given/known data
I am not sure where I am having trouble here.

A car id driven 125km west and then 95km southwest. What is the mag and direction of displacement?

2. Relevant equations
Triangle method says that if I draw the vectors from tip to tail, the displacement vector is the vector that connects the tail of the first vector to the head of the last one.

3. The attempt at a solution I am assuming (hopefully correctly!) that magnitude is just 230km, but I cannot sem to find direction.

If it were 125 west and 95 south of west I could figure it out, but it is 125 west and 95 southwest....so I can't use arctan can I?

2. Sep 30, 2007

### alandry06

Wait...after looking at it again, I think the magnitude is wrong too...

3. Sep 30, 2007

### Dick

It probably isn't correct. Split both displacements into west and south components using sin and cos. Then add the west and south components separately. Finally, you use arctan.

4. Sep 30, 2007

### alandry06

I thought of that but I am not given any angles?

5. Sep 30, 2007

### Dick

95km southwest is 95*sin(45)km west plus 95*sin(45)km south. Does that help? Your final displacement should be mostly west and some south. How much of each? So for an angle you could just say how many degrees south of west is it.

6. Sep 30, 2007

### alandry06

Wait...is it safe to assume that when they say southwest they mean exactly 45 degrees between west and south?.....I am going to bank on that being the case.

Edit:Thanks Dick! It took me awhile to type that...but it appears you confirmed my idea. I should be able to get this from here.

7. Sep 30, 2007

### alandry06

So (95sin45)^2+(95cos45+125)^2=mag and the angle would be
arctan (abs value[(95sin45)/(125+95cos45)]) south of west.

8. Sep 30, 2007

### Dick

Right. Except that's mag^2. Don't forget the sqrt.