A boat has a mass of 4040 kg. Its engines generate a drive force of 4660 N due west, while the wind exerts a force of 880 N due east and the water exerts a resistive force of 1400 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign?

Answer: acceleration = 0.59m/s2

Direction is due west

Explanation: M = 4040

F1 = 4660

F2=1400

F3=880

Net force = F1 - (F2 + F3)

= 4660 - (1400 + 880)

=2380

Acceleration (a) = F/M

= 2380/4040

= 0.59m/s2

Given:

Mass of the boat, m = 4040 kg

The driving force of engine, FB = 4660 N in west = + 4660 N

The force of wind, Fwi = 880 N in east = - 880 N

The force of water, Fwa = 1400 N in east = - 1400N

Total three forces are acting on the boat

Fnet = Fb+fwi+Fwa

Fnet = 4660 - 880 - 1400

Fnet = + 2380N

Acceleration (a) = Force/mass

= 2380/4040

= 0.59m/s2