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 10° north of east.

What is his final position relative to the island?
______ km
______ ° south of east

Answer 1

Suppose given that Vector is R and it makes angle A with +x-axis, then it's components are given by:

Rx = R*cos A

Ry = R*sin A

Using above rule:

d1 = 2.5 km 45° north of west = 2.5 km 135 deg CCW from +ve x-axis

d1x = 2.5*cos 135 deg = -1.77 km

d1y = 2.5*sin 135 deg = 1.77 km

d2 = 4.70 km 60° south of east = 4.70 km 330 deg CW from +ve x-axis

d2x = 4.7*cos (300 deg) = 2.35 km

d2y = 4.7*sin (300 deg) = -4.07 km

d3 = 5.1 km straight east = 5.1 km on +ve x-axis

d3x = 5.1*cos 0 deg = 5.1 km

d3y = 5.1*sin 0 deg = 0 km

d4 = 7.2 km 55° south of east = 7.2 km 235 deg CCW from +ve x-axis

d4x = 7.2*cos (235 deg) = -4.13 km

d4y = 7.2*sin (235 deg) = -5.90 km

d5 = 2.8 km 10° north of east = 2.8 km 10 deg CCW from +ve x-axis

d5x = 2.8*cos (10 deg) = 2.76 km

d5y = 2.8*sin (10 deg) = 0.49 km

So net displacement will be

D = (d1x + d2x + d3x + d4x + d5x) i + (d1y + d2y + d3y + d4y + d5y) j

D = (-1.77 + 2.35 + 5.1 - 4.13 + 2.76) i + (1.77 - 4.07 + 0 - 5.90 + 0.49) j

D = 4.31 i - 7.71 j

Magnitude of above position will be

|D| = sqrt (4.31^2 + (-7.71)^2)

|D| = 8.83 km

Direction of above position will be

direction = arctan (-7.71/4.31) = -60.8 deg

direction = 60.8 deg south of east

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