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 45.0∘ north of west; then 4.70 km 60.0∘ south of east; then 1.30 km 25.0∘ south of west; then 5.10 km straight east; then 1.70 km 5.00∘ east of north; then 7.20 km 55.0∘ south of west; and finally 2.80 km 10.0∘ north of east. What is his final position relative to the island?

The position is (3.06i-6.34j) = 7.04 Km 64.2 south of east

Explanation:

We can solve this problem by two methods one analytical, using trigonometry and another graph, let's start with the analytical method, we start by decomposing each position vector

V1 = 2.50 Km 45.0º norte del oeste

V1x = V1 cos 

V1y = Vi sin 

All angles must be measured from the positive side of the x axis

45.0º norte del oeste = 180º - 45 = 135º

V1x = 2.50 cos 135

V1y = 2.50 sin 135

V1x = - 1.77 Km

V1y = 1.77 Km

We repeat this same procedure for all vectors, now we are going to find the angles, let's not forget that all the angles are measured with respect to the positive part of the x-axis counterclockwise

V2 = 4.70 Km 60.0º south of east

60.0º south of east = 360 - 60 = 300º

V2x = 4.7 cos 300 = 2.35 Km

V2y=4.7 sin 300 = - 4.07 Km

V3 = 1.30 Km 25.0º south of west

25.0º south of west = 180 + 25 = 205º

V3x = 1.3 cos 205 = - 1.17 Km

V3y = 1.3 sin 205 = - 0.55 Km

V4 = 5.1 Km straight east

East = 0º

V4x = 5.1 Km

V4y=0 Km

V5 = 1.7 Km 5.00º east of north

5.00º east of north = 5º

V5x = 1.7 cos 5 = 1.69 Km

V5y = 1.7 sin 5 = 0.15 Km

V6 = 7.20 Km 55.0º south of west

55.0º south of west = 270-55 = 215º

V6x = 7.2 cos 215=-5.90 Km

V6y = 7.2 sin 215 = - 4.13 Km

V7 = 2.80 Km 10.0º north of east

10.0º north of east = 10º

V7x = 2.80 cos 10 = 2.76 Km

V7y = 2.80 sin 10 = 0.49 Km

Now we can make the sum for each axis

Vtx = V1x+V2x+V3x+V4x+V5x+V6x+V7x

Vty = V1y+V2y+V3y+V4y+V5y+V6y+V7y

Vtx = - 1.77+2.35-1.17+5.1+1.69-5.90+2.76

Vtx = 3.06 Km

Vty = 1.77-4.07-0.55+0+0.15-4 ... 13+0.49

Vty = - 6.34 Km

The answer is (3.06i-6.34j), it can also be given as a module and angle

Vt2 = Vtx2 + Vty2

θ = tan-1 (Vty/Vtx)

Vt = 3.062 + (-6.34) 2

θ = tan-1 (-6.34/3.06)

Vt = 7.04 Km

θ = - 64.2º

-64.2º = 64.2 south of east