Answer:
a = 4.04 m/s2
[tex] 53.44^o [/tex]north of east
Explanation:
The mass of the boat is m = 1200 kg
The forces acting due to wind on the boat is Fwind = 3900 N j^ (north)
and The forces acting due to water on the boat is Fwater = 2900 N i^ (east)
The net force on the boat is F = 2900 N i^ + 3900 N j^
The horizontal component of the net acceleration is ax = Fwater/m = 2.41 m/s2
vertical component of the net acceleration is ay = Fwind/m = 3.25 m/s2
Net acceleration is [tex]a = \sqrt{[(2.41)^2 + (3.25)^2]}[/tex]
a = 4.04 m/s2
The direction of the net acceleration is [tex]\theta = tan^{-1}\frac{3.25}{2.41} [/tex]=
=[tex] 53.44^o [/tex]north of east
