To sail directly north at 32 knots

Question

Rear Admiral Crunch wants his yacht to sail directly north at a speed of 32 knots. What course and speed should he tell his helmsman to make if there is an ocean current going in the direction of 65° to the west of north at a rate of 8 knots?

I drew an image and all I know that when he wants to change the course of this journey, he needs to cancel out the x reaction force and also add extra y reaction force.

Answer 1

This is simple vector subtraction. Breaking the ocean current vector into north/south and east/west components we find that it is the sum of

• a vector pointing due west with magnitude $8\cos25^\circ$ knots
• a vector pointing due north with magnitude $8\sin25^\circ$ knots

Then it is clear that to give an overall velocity of 32 knots due north, the yacht must be going at the sum of

• a vector pointing due east with magnitude $8\cos25^\circ$ knots, to cancel the first vector above
• a vector pointing due north with magnitude $32-8\sin25^\circ$ knots

Calculating the sum of these two vectors yields $$r=\sqrt{(8\cos25^\circ)^2+(32-8\sin25^\circ)^2}=29.523$$ $$\theta=\tan^{-1}\frac{8\cos25^\circ}{32-8\sin25^\circ}=14.216^\circ$$ Therefore the yacht should be going at 29.5 knots at 14.2° east of north.