having trouble with this one
The vector sum of all these vectors would be your answer.
But you need to express every vector in the form, $\displaystyle <a,b>$---> The components of your vector.
---
In order to solve this problem we need to develope a formula. Let $\displaystyle \theta$ be the angle from the x-axis (meaning measure conterclockwise).
Then if $\displaystyle (x,y)$ are the endpoints of the vector. The length is $\displaystyle l=\sqrt{x^2+y^2}$. And$\displaystyle \sin \theta = \frac{y}{\sqrt{x^2+y^2}}$
In order to solve this problem. You need to solve for both $\displaystyle x,y$
More simply,
$\displaystyle \sqrt{x^2+y^2}=l$
$\displaystyle \sqrt{x^2+y^2}\sin \theta =y$
Now, divide the second equation by the first to get, (it is nonzero)
$\displaystyle \sin \theta = \frac{y}{l}$
Thus,
$\displaystyle l\sin \theta = y$
Similary,
$\displaystyle l\cos \theta = x$
As a result we have converstion formula:
If $\displaystyle x,y$ the the endpoints of a vector having length $\displaystyle l$ and angle meausured from the x-axis $\displaystyle \theta$ then,
$\displaystyle \left\{ \begin{array}{c} l\sin \theta =y\\l\cos \theta = x \end{array}\right\}$
---
Now construct a chart in terms of $\displaystyle \theta$, i.e. counterclockwise angle.
$\displaystyle \left\{ \begin{array}{cc}\mbox{Leg #} & \theta\\
1&135^o\\2&345^o\\3&215^o\\4&90^o\\5&0^o\\6&264^ o $
Therefore you new table is,
$\displaystyle \left\{ \begin{array}{ccc}\mbox{Leg #}&\mbox{length}&\theta\\
1&22.5&135^o\\
2&35.4&345^o\\
3&18.6&215^o\\
4&15.5&90^o\\
5&31.8&0^o\\
6&44.8&264^o $
Now use the converstions formula posted above to find the coordinates.
So you have,
$\displaystyle \left\{ \begin{array}{cc}x&y\\
15.91&-15.91\\
-9.16&34.19\\
-10.67&-15.24\\
15.5&0\\
0&31.8\\
-44.55&-4.68$
Now by the rule of vectors you can add their components thus,
$\displaystyle \left\{ \begin{array}{cc}\sum x&\sum y\\
-32.97&30.16
$
Now compute the length,
$\displaystyle \sqrt{(32.97)^2+(30.16)^2}\approx 44.68\mbox{ km}$
Hello,Originally Posted by bigstarz
in addition to TPHs result: Calculate the direction between starting point to ending point by the components of the resulting vector [-32.97 , 30.16]:
$\displaystyle \sin(\theta)=\frac{-32.97}{30.16}$. Thus $\displaystyle \theta\approx 113.8^o$
Greetings
EB