# 1 vector question

1. Oct 7, 2005

### kevinlikesphysics

5. [SFHS99 3.P.29.] A person walks the path shown in Figure 3-27. The total trip consists of four straight-line paths. At the end of the walk, what is the person's resultant displacement measured from the starting point? Let L = 310.0 m, and let = 50.0°

Magnitude____ m
Direction____________° (counterclockwise from the person's initial direction)

the picture is in the link below
my answer was 336.7 but i was wrong i knew it because it was too big anyway but i dont know how to get it can someone explain

http://www.webassign.net/sfhs99/3-27alt.gif

2. Oct 7, 2005

### arildno

Hi, Kevin; it would be best if you post the vectors you tried to use in deriving your answer; I suspect you've just made a sign error or something like that.

3. Oct 7, 2005

### kevinlikesphysics

df

i pasted the image what do you mean show my vectors?

i jsut split each vector on the image into x and y components for each and added them to found the resulting vector is that what i was suposed to do

4. Oct 7, 2005

### mezarashi

Could you type out those calculations with the x and y components? This is pretty straight forward. I'll be looking to catch a arithmetic mistake in your work.

5. Oct 7, 2005

### arildno

What I meant is quite simple:
What vectors did you actually use?

Let $$\vec{v}_{1}=100(1,0), \vec{v}_{2}=L(0,-1), v_{3}=150(-\cos(30),-\sin(30)), \vec{v}_{4}=200(-\cos(\theta),\sin(\theta))$$

the displacement vector you're seeking, is simply the vectorial sum of these.

6. Oct 7, 2005

### kevinlikesphysics

x y

100 0

0 -310

-75 -129.9

-153.2 128.56
_____________________
-128.2 -311.34

7. Oct 7, 2005

### arildno

You have your values mixed up. The sine to 30 degrees is 1/2, that is 75 is the y-component, not the x-component.
In addition, the cosine to 50 degrees must be less than the sine to 50 degrees, so the values in your 4th vector is definitely mixed up as well.

8. Oct 7, 2005

### kevinlikesphysics

is it

x total = -183.1 and y total = -558.2

edit: i got it right .. thanks for the help

R = 280.78

angle 235.64323239883773379400378289096 degrees cc

Last edited: Oct 7, 2005