Question

The force exerted by the wind on the sails of a sailboat is fsail = 410 n north. the water exerts a force of fkeel = 200 n east. if the boat (including its crew) has a mass of 300 kg, what are the magnitude and direction of its acceleration?

Answer 1

The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse.
(200)² + (410)² = F²
40,000 + 168,100 = F²
√(208,100) = F
456.18 N = F

F= ma
456.18 N = (300 kg)a
456.18 / 300kg = a
1.52 m/s² = a

The sailboat will be heading North East. To  find the angle of the boats trajectory use inverse tangent function.
tan(Ф) = opposite/adjacent
arctan(opposite/adjacent) = Ф

arctan(410/200) = 64° North East