Answer:
R=7.34km ∠ 63.47° South of East.
Explanation:
We can either solve this graphically or by using the components of the given vectors. I'll solve it by using the components. So first we need to do a sketch of what the displacements will look like (See attached picture).
The piture is not drawn to scale, but it should help us visualize the directions better. I will name each displacement a different letter so we can distinguish them. We will say that if he goes north, he will have positive y-displacement, if he goes south, he will have negative y-displacement, if he goes east, he will have positive x-displacement and if he goes west, he will have negative x-displacement.
So having said this let's begin. So let's take the first displacement, displacement A:
2.50 km 45° north of west.
He goes north here which means he has positive y-displacement and he goes west, which means he has negative x-displacement. Its components can be found by using the sin and cos functions, like this:
[tex]A_{x}=Acos(\theta)[/tex]
[tex]A_{x}=2.50 cos (45^{o})[/tex]
[tex]A_{x}=-1.7678km i [/tex]
[tex]A_{y}=Asin(\theta)[/tex]
[tex]A_{y}=2.50 sin (45^{o})[/tex]
[tex]A_{y}=1.7678km j [/tex]
and we do the same with the other vectors.
B = 4.70km 60.0° south of east:
[tex]B_{x}=2.35i km[/tex]
[tex]B_{y}=-4.070j km[/tex]
C = 1.30km 25.0° south of west:
[tex]C_{x}=-1.1782i km[/tex]
[tex]C_{y}=-0.549j km[/tex]
D = 5.10km straight east:
[tex]D_{x}=5.10i km[/tex]
[tex]D_{y}=0j km[/tex]
E = 1.70km 5.00° east of north
In this case we need to flip the functions due to the direction of the displacement.
[tex]E_{x}=Esin(\theta)[/tex]
[tex]E_{x}=1.70 sin (5^{o})[/tex]
[tex]E_{x}=0.1482km i [/tex]
[tex]E_{y}=Ecos(\theta)[/tex]
[tex]E_{y}=1.70 cos (5^{o})[/tex]
[tex]E_{y}=1.6935km j [/tex]
F = 7.20km 55.0° south of west:
[tex]F_{x}=-4.13i km[/tex]
[tex]F_{y}=-5.898j km[/tex]
G = 2.80km 10.0° north of east:
[tex]G_{x}=2.757i km[/tex]
[tex]G_{y}=0.486j km[/tex]
So now it's time to find the resulting vector, so we get:
[tex]R_{x}=A_{x}+B_{x}+C_{x}+D_{x}+E_{x}+F_{x}+G_{x}[/tex]
[tex]R_{x}=-1.7678i+2.35i-1.1782i+5.10i+0.1482i-4.13i+2.757i[/tex]
[tex]R_{x}=3.2792i km[/tex]
[tex]R_{y}=A_{y}+B_{y}+C_{y}+D_{y}+E_{y}+F_{y}+G_{y}[/tex]
[tex]R_{y}=1.7678j-4.070j-0.549j+0j+1.6935j-5.898j+0.486j[/tex]
[tex]R_{y}=-6.5697j km[/tex]
So, now that we have the components of the resultant vector we can find magnitude of the displacement and the angle.
[tex] |R|=\sqrt {(3.2792)^{2}+(-6.5697)^{2}} [/tex]
which yields
|R|=7.34 km
∠ = [tex] tan^{-1}(\frac{-6.5697}{3.2792}) [/tex]
∠ = -63.47°
So the final position relative to the island is:
7.34km 63.47° South of east
