Question

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 2.5 km 45° north of west; then 4.70 km 60° south of east; then 5.1 km straight east; then 7.2 km 55 south of west; and finally 2.8 km 5 north of east. What is his final position in kilometers and degrees south of east relative to the island?

Answer 1

Answer:8.28 km

Explanation:  

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$=$\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$+$\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

for direction

$tan\theta =\frac{7.95}{2.299}=3.4580$

$\theta =73.87^{\circ}$ south of east

Answer 2

Explanation:

Let us assume the direction east as i, direction west as -i, direction north as j and direction south as j.

Now, we define each of the straight lines as a vector with components along each of these unit vectors, we get

A : 2.5 km 45° north of west

A = 2.5 cos45 (-i) + 2.5 cos45 (j)

  = -1.77(i) + 1.77(j)

B : 4.7 km 60° south of east

B = 4.7 cos60 (i) + 4.7 sin60 (-j)

  = 2.35(i) - 4.07(j)

C : 5.1 km straight east

C = 5.1(i)

D : 7.2 km 55° south of west

D = 7.2 cos55 (-i) + 7.2 sin55 (-j)

   = -4.13(i) - 5.9(j)

E: 2.8 km 5° north of east

E = 2.8 cos5 (i) + 2.8 sin5 (j)

   = 2.79(i) + 0.24(j)

The resultant sum of all these vectors is  as follows.

                   R = A + B + C + D + E

The sum of all the above yields,

R = 4.34(i) + 7.96(-j)

The magnitude of the resultant is  as follows.

$|R| = \sqrt{4.34^{2} + (-7.96)^{2}}$

         = 9.07

The angle that the resultant makes is  as follows.

$\theta = tan^{-1}(\frac{7.96}{4.34}) = 61.4^{o}$

So, his final position relative to the island is  as follows.

9.07 km, 61.4° south of east.

Thus, we can conclude that final position of Gilligan in kilometers and degrees south of east relative to the island is  9.07 km, 61.4° south of east.