Answer :
Answer:
[tex]v_{wE}=2.334m/s[/tex]
[tex]\alpha_{wE}=44.81^\circ[/tex]
Explanation:
The velocity of the wind relative to the Earth is the sum of the velocity of the wind relative to the sailboat and the velocity of the sailboat relative to the Earth, [tex]v_{wE}=v_{ws}+v_{sE}[/tex], as vectors. We always need to separate into components, and as usual we will take East as the positive x direction, North as the positive y direction, and the angles [tex]\theta[/tex] measured from the East direction and anticlockwise.
The velocity of the sailboat relative to the Earth [tex]v_{sE}[/tex] is the velocity of the ocean current:
[tex]v_{sEx}=v_{sE}cos(\theta_{sE})=(1.86m/s)cos(90^\circ-25.1^\circ)=0.789m/s[/tex]
[tex]v_{sEy}=v_{sE}sin(\theta_{sE})=(1.86m/s)cos(90^\circ-25.1^\circ)=1.684m/s[/tex]
We do the same for the velocity of the wind relative to the sailboat [tex]v_{ws}[/tex]:
[tex]v_{wsx}=v_{ws}cos(\theta_{ws})=(4.13m/s)cos(180^\circ+53.7^\circ)=-2.445m/s[/tex]
[tex]v_{wsy}=v_{ws}sin(\theta_{ws})=(4.13m/s)cos(180^\circ+53.7^\circ)=-3.329m/s[/tex]
And finally use the original formula for each direction:
[tex]v_{wEx}=v_{wsx}+v_{sEx}=0.789m/s-2.445m/s=-1.656m/s[/tex]
[tex]v_{wEy}=v_{wsy}+v_{sEy}=1.684m/s-3.329m/s=-1.645m/s[/tex]
With the components we can easily calculate the magnitude of the velocity vector:
[tex]v_{wE}=\sqrt{v_{wEx}^2+v_{wEy}^2}=2.334m/s[/tex]
For the angle, we must notice that the components are negative, so by taking [tex]v_{wEy}[/tex] as the opposite cathetus and [tex]v_{wEx}[/tex] as the adjacent one we will have our angle in degrees south of west (which for avoiding confusion with the previous convention I'll call [tex]\alpha_{wE}[/tex]):
[tex]\alpha_{wE}=Arctg (\frac{v_{wEy}}{v_{wEx}})=44.81^\circ[/tex]