A canoe has a velocity of 0.49 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.45 m/s east relative to the earth.
A) Find the magnitude of the velocity of the canoe relative to the river.
B) Find the direction of the velocity of the canoe relative to the river – express your answer as an angle measured south of west.
I’m having some trouble with this, please explain how you got your answer, thanks!
The velocity of the canoe relative to the river is equal to the velocity of the canoe minus the velocity of the river, use vector addition/subtraction…
i^ = the unit vector in the x-direction, setting east as positive
j^ = the unit vector in the y-direction, setting north as positive
Vc = 0.49cos(45°) i^ + -0.49sin(45°) j^
Vr = 0.45 i^
Vc/r = Vc – Vr = [0.49cos(45°) – 0.45] i^ + [-0.49sin(45°) – 0] j^
Vc/r = -0.1035 i^ + -0.3465 j^
Magnitude Vc/r = √((-0.1035)² + (-0.3465)²) = 0.3616 m/s <===============
Angle = arctan (-0.3465 / -0.1035) = 73.4° south of west <===============
Perfect explanation, short and simple! Thank you!!