Question

23. In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30km 25.0º south of west; then 5.10 km straight east; then 1.70km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?

Answer 1

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:

$A_{x}=Acos(\theta)$

$A_{x}=2.50 cos (45^{o})$

$A_{x}=-1.7678km i$

$A_{y}=Asin(\theta)$

$A_{y}=2.50 sin (45^{o})$

$A_{y}=1.7678km j$

and we do the same with the other vectors.

B = 4.70km 60.0° south of east:

$B_{x}=2.35i km$

$B_{y}=-4.070j km$

C = 1.30km 25.0° south of west:

$C_{x}=-1.1782i km$

$C_{y}=-0.549j km$

D = 5.10km straight east:

$D_{x}=5.10i km$

$D_{y}=0j km$

E = 1.70km 5.00° east of north

In this case we need to flip the functions due to the direction of the displacement.

$E_{x}=Esin(\theta)$

$E_{x}=1.70 sin (5^{o})$

$E_{x}=0.1482km i$

$E_{y}=Ecos(\theta)$

$E_{y}=1.70 cos (5^{o})$

$E_{y}=1.6935km j$

F = 7.20km 55.0° south of west:

$F_{x}=-4.13i km$

$F_{y}=-5.898j km$

G = 2.80km 10.0° north of east:

$G_{x}=2.757i km$

$G_{y}=0.486j km$

So now it's time to find the resulting vector, so we get:

$R_{x}=A_{x}+B_{x}+C_{x}+D_{x}+E_{x}+F_{x}+G_{x}$

$R_{x}=-1.7678i+2.35i-1.1782i+5.10i+0.1482i-4.13i+2.757i$

$R_{x}=3.2792i km$

$R_{y}=A_{y}+B_{y}+C_{y}+D_{y}+E_{y}+F_{y}+G_{y}$

$R_{y}=1.7678j-4.070j-0.549j+0j+1.6935j-5.898j+0.486j$

$R_{y}=-6.5697j km$

So, now that we have the components of the resultant vector we can find magnitude of the displacement and the angle.

$|R|=\sqrt {(3.2792)^{2}+(-6.5697)^{2}}$

which yields

|R|=7.34 km

∠ = $tan^{-1}(\frac{-6.5697}{3.2792})$

∠ = -63.47°

So the final position relative to the island is:

7.34km 63.47° South of east