You want your boat to travel due east at 20 mph. If the wind is blowing at 9 mph with a direction 30 degrees south of west, to what direction and speed should you set your course?

Question

From my diagram, I can see that the components of the wind vector is 9<cos(210), sin(210)> and my boat vector is 20<cos(0), sin(0)>. However, when I add the components and find the magnitude, it gives a value of 13, which seems way too small.

When I used the cosine and sine rule, I found that the magnitude was 28, which makes a bit more sense. However, I would like to solve it using vector components if possible.

Answer 1

You want to solve the vector equation

<x,y> + 9<-√3/2, -1/2> = <20,0>

x = 20+9√3/2

y = 0+4.5

Which gives

x = 27.794

y = 4.5

<27.794, 4.5>

or

28.156 ∠9.197º

Answer 2

The course vector is unknown with coordinates (rcosθ,rsinθ).

The resultant vector is the desired outcome (20,0).

So,

rcosθ+9cos210 = 20

rsinθ+9sin210=0

Two equations, two unknowns. From the 2nd equation, r = (-9sin210)/sinθ. Substitute that expression into the first equation for r and solve for θ. Once you have θ, plug into equation where you have r in terms of θ.

desmos.com/calculator/jaj2kzkkls

Answer 3

You should not be adding these vectors! V_B + V_W = V where the first is the velocity of the boat with respect to still air the second is the velocity of the air with respect to the earth, and the last velocity is the velocity of the boat with respect to the earth.

You know V and V_W You can solve for V_B by adding V and -V_W ( which is 9 mph in the 30° direction. )

You can do this in x and y (by components) and then resolve the V by V_B = sqrt(V_Bx²+V_By²) and θ_B = arctan(V_y/V_x) with θ = 180º + θ for V_x<0

The x component looks like V_Bx = Vcos(0º) + V_Wcos(30º)

y is same with sines