The force exerted by the wind on the sails of a sailboat is 360 N north. The water exerts a force of 170 N east. If the boat (including its crew) has a mass of 310 kg, what are the magnitude and direction of its acceleration?

The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse. Using trigonometry equations,

(170) ² + (360) ² = F²

28900 + 129600 = F²

√ (158500) = F

398.12 N = F

Remember,

F = ma

398.12 N = (310 kg) * a

398.12/310kg = a

1.28 m/s² = a

The sailboat will be heading North East. The angle of the boats trajectory use inverse tangent function.

tan (Ф) = opposite/adjacent

= arctan (opposite/adjacent)

= arctan (360/170)

= 64.7° North East.

The magnitude and direction of its acceleration is 1.284 m/s² North-East

Explanation:

Given;

The force exerted by the wind on the sails of a sailboat to be 360 N north

The force exerted by water on the sails of a sailboat to be 170 N east.

The resultant force on the boat is can be calculated by Pythagoras theorem, if these two forces makes right angled-triangle.

R² = N² + E²

R² = (360) ² + (170) ²

R² = 158500

R = √ (158500)

R = 398.121 N

If the If the boat (including its crew) has a mass of 310 kg, then the magnitude and direction of its acceleration becomes;

a = F/m

a = 398.121/310

a = 1.284 m/s² North-East