Answer :
Answer:
a) Figure attached
b) [tex] v_r = \sqrt{3.55^2 +10.94^2}=11.50 mph[/tex]
[tex] \theta = tan^{-1} (\frac{10.94}{3.55})=72.02[/tex]
Step-by-step explanation:
Part a
See the figure attached.
Part b
For this case first we need to find the vectors of velocity for the boat and the wind like this:
[tex] b= <10 cos(65), 10 sin(65)>= <4.23, 9.06> mph[/tex]
[tex] w= <2 cos(110), 2 sin(110)>= <-0.68, 1.88> mph[/tex]
And now if we want to find the resulting velocity we just need to add the vector:
[tex] b + w = <4.23-0.68, 9.06+1.88>=<3.55, 10.94>mph[/tex]
And the resultant magnitude would be:
[tex] v_r = \sqrt{3.55^2 +10.94^2}=11.50 mph[/tex]
And if we want the resultant angle we can do this:
[tex] \theta = tan^{-1} (\frac{10.94}{3.55})=72.02[/tex]
