Question

a sailboat sails 50 km to the north then 80 km in the direction 38 degrees west of south. what is the net displacement of the boat.

Answer 1

Answer:

35.489km

Explanation:

a diagram illustrating the question is attached

now, applying cosine rule to find the displacement c

cosine rule;

c²=a²+b²-2abCos ∅

c²=50²+80²-2(50)(80)Cos 38°

c²=2500 + 6400 - (8000×0.9951)

c²=8900-7640.589

c²=1259.411

c=$\sqrt{1259.411}$

c=35.49km

Answer 2

Answer:

S = 49.25i + 113.04j with magnitude 123.22km

Explanation:

The net displacement has both x and y components.

S = S(x)i + S(y)j

The x component of the net displacement is given as the sum of the x component of the first and second displacements:

S(x) = S(1,x) + S(2,x)

S(1,x) = 50 cos(90)

S(1,x) = 0 km

S(2,x) = 80 sin (38)

S(2,x) = 49.25km

=> S(x) = 0 + 49.25

S(x) = 49.25km

The y component of the net displacement is given as the sum of the y component of the first and second displacements:

S(y) = S(1,y) + S(2,y)

S(1,y) = 50 sin(90)

S(1,y) = 50km

S(2,y) = 80 cos (38)

S(2,y) = 63.04 km

=> S(y) = 50 + 63.04

S(y) = 113.04 km

=> S = 49.25i + 113.04j

Hence, the magnitude of the net displacement will be

S = √(S(x)² + S(y)²)

S = √(49.25² + 113.04²)

S = √(15183.94)

S = 123.22km