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)

I’m good with pshycology

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well