Vectors Question

Question

A sailboat race course consists of four legs, defined by the displacement vectors A, B, C, and D, as the drawing indicates. The magnitudes of the first three vectors are A = 3.40 km, B = 5.10 km, and C = 5.30 km. The finish line of the course coincides with the starting line. Using the data in the drawing, find the distance of the fourth leg and the angle θ.

 

http://www.webassign.net/CJ/1-49.gif

Answer 1

Since the race finishes at the starting location the net displacement vector will be zero.

 

A + B + C + D = 0

 

Solve for the components of vector D taking up and right as positive.

 

Horizontal component (right is positive, left is negative)

 

3.40cos(40) - 5.10cos(35) - 5.30cos(23) + Dcos(θ) = 0 

-6.45 + Dcos(θ) = 0

Dcos(θ) = 6.45 km

 

 

Vertical Component (up is positive, down is negative)

 

3.40sin(40) + 5.10sin(35) - 5.30sin(23) - Dsin(θ) = 0

3.04 - Dsin(θ) = 0

3.04 km = Dsin(θ)

 

Trig Identity

tan(θ) = sin(θ)/cos(θ)

tan(θ) = 3.04/6.45

arctan(3.04/6.45) = θ

25.24° = θ

 

Plug in the angle

 

Dcos(25.24) = 6.45 km

D = 6.45/cos(25.24)

D = 7.13 km