Resultant Displacement: L=310m, θ=50°

[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 don't know how to get it can someone explain



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

 

[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.

 

[kevinlikesphysics]

df

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

i just 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

 

[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.

 

[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.

 

[kevinlikesphysics]

x y

100 0

0 -310

-75 -129.9

-153.2 128.56
_____________________
-128.2 -311.34

 

[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.

 

[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