Answer:
(a) θ = 55.85 degree
(b) 7.89 km
Explanation:
Using vector notations
A = 1.88 km south = 1.88 (- j) km = - 1.88 j km
B = 9.05 km 47 degree north of east
B = 9.05 ( Cos 47 i + Sin 47 j) km
B = (6.17 i + 6.62 j) km
Net displacement is
D = A + B
D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j
(a) Angle made with positive X axis
tanθ = 6.62 / 4.29 = 1.474
θ = 55.85 degree
(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]
distance = 7.89 km