If a 990 kg car is traveling on the road and the Ff is 360 N to the east and the applied force is 1330 N to the west, is the car displaying acceleration or in constant motion? Choose the best answer to describe the car's motion. (Remember to draw a force diagram to solve).

The force diagram has four forces: two in the horizontal direction (west - east) and two in the vertical direction (up and down).

The two forces in the horizontal direction are:

1) Force of friction, Ff = 360 N east.

2) Force applied = 1330 N west

=> Net horizontal force = 1330N west - 360N east = 970 west.

Vertical forces:

3) Weight = 990 kg * 9.8 m/s^2 = 9,702 N downward

4) Normal force = - weight = - 9,702N upward

Net vertical force = 9,702N downward - 9,702N upward = 0

5) Second Law of Newton

Net force = m * a = > a = Net force / m = 970 N west / 990 kg = 0.98 m/s^2

Then, the conclusion is that the car is moving with uniform acceleration of 0.98 m/s^2 westward (that magnitud is 10% of gravitational acceleration)