A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point.

Question

a) Find the magnitude of the third leg of the journey.

b) Find the direction of the third leg of the journey. (Answer = East of North)

Anyone mind helping please.

Answer 1

Break the 3 vectors into components and solve for the unknown vector.
For this choose
North is positive y
East is positive x

First vector, 2.00km east represented by (2.00km)x + 0(km)y

Second vector, 3.50km SE represented by (3.50)(cos(315))(km)x + (3.50)(sin(315))(km)y

Third vector is unknown, represent it by (a)(km)x + (b)(km)y.

All three vectors added = 5.80(km)x + 0y
so:
2.00(km)x + 3.50(cos(315))(km)x + a(km)x = 5.80(km)x
Solve for a

0(km)y + 3.50(sin(315))(km)y + b(km)y = 0(km)y
Solve for b.
The resulting angle with this method will be relative to the x axis. Adjust the answer to reflect the degrees from the y axis.

Answer 2

why is it not 45 degress why 315

Answer 3

Here the vector angles are relative to the X axis. So, east would be 0 degrees, north is 90 degrees, west is 180 degree, and south is 270 degrees. So, SE is 270 + 45 = 315. -45 degrees could also be used.

Answer 4

45

Answer 5

Dame desu

Answer 6

129485