Question

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day and he is blown along the following directions: 2.80 km and 45.0° north of west, then 4.20 km and 60.0° south of east, then 1.50 km and 25.0° south of west, then 5.10 km straight east, then 1.70 km and 6.00° east of north, then 7.20 km and 55.0° south of west, and finally 2.80 km and 10.0° north of east. Use a graphical method to find the castaway's final position relative to the island. (Enter the magnitude in kilometers and the direction in degrees south of east.)

Answer 1

Convert all value to x and y Cartesian measures where +x is north and + y is east

If you measure your angle ccw of due east

x = r*cos(theta)

y = r*sin(theta )

holds true

so convert all angles to ccw from due east

(1) 2.8 km 45° north of west; then

45 North of west is 180-45 = 135 degrees from due east

x= 2.8 * cos(135)

y = 2.8* sin(135)

(2) 4.20 km 60° south of east; then

60 south of east is -60 degrees counterclockwise for east

x = 4.2 *cos( -60)

y = 4.2 * sin(-60)

(3) 1.5 km at 25 degrees south of west, which is 180+ 25 ccw of due east

x = 1.5*cos(205)

y = 1.5*sin(205)

(4) 5.1 km straight east; then

x = 5.1 km

y = 0

(5) 1.7 km 6° east of north; then

6 east of north is 90-6 = 84 degrees counterclockwise for east

x = 1.7 *cos( 84)

y = 1.7 * sin(84)

(6) 7.20 km 55° south of west; then

55 south of west is 180+ 55=235 ccw of due east

x = 7.2 *cos(235)

y = 7.2 * sin(235)

(7) 2.8 km 10° north of east.

10 degrees north of east of east is 10 ccw of due east

x = 2.8*cos(10)

y = 2.8*sin(10)

so add x's up and add all y's up, we get:

x = 2.666 km

y = -6.01 km; i.e. 6.01 km south

Magnitude = sqrt(2.666^2 + 6.01^2) = 6.58 km

Direction = arctan (6.01/2.666) = 66.1 degrees south of east