I will do this one using vectors.
I will also use the convention: due S.W ----> 225°, W.15.N ----> 165°
vector R = (12cos225°, 12sin225°) + (4cos165°, 4sin165°
= (-8.485..., -8.485...) + (-3.863..., 1.0353..) = (-12.348..., -7.45...)
|R| = √((-12.348..)^2 + (-7.45..)^2 ) = appr 14.4
So the velocity is 14.4 km/h
direction: tanØ = -7.45/-12.348 , Ø = (180+31.1)° = appr 211.1° , or using your notation W 31.1° S
I am sure you can find the answer to the last part of your problem
A ship that can cruise at 12km/h in still waters , sets course due S.W .it is driven off by current flowing W.15.N at a rate of 4 km/h.calulate : a. The resultant velocity
b. The displacement after 3 hours.
1 answer
1 answer