Posted by Joe Rogers on Saturday, September 29, 2012 at 3:34pm.
Whoever wrote this question knows nothing about the physics of sailing.
The tidal current does not push the hull in general unless the boat is anchored to something. The current is just a velocity component of the boat. It moves with the water at the speed of the water.
All I can do is guess that there are only the two forces given in the problem acting on the boat and calculate the magnitude and direction of the acceleration.
north component of force = 55.2^103 cos 45 = 39*10^3 N north
east component of force = 16.4*10^3 - 39*10^3 = -22.6*10^3 Newtons East
magnitude of force = 10^3 sqrt(39^2+22.6^2)
= 45*10^3 Newtons
a = F/m = 45*10^3/19*10^3 = 2.37 agreed
tan of angle north of west = 39/22.6
angle north of west = 60 degrees north of west or 30 degrees west of north or 330 degrees on a compass clockwise from north.
By the way the f=ma has the wrong m if we were really doing a boat in water. When you accelerate a boat in water, you also accelerate some water around it leading to what is called "added mass" in hydrodynamics. As I said, whoever wrote this question knows nothing about the physics of boats.
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