Question

A 1502.7 kg car is traveling at 33.1 m/s when

the driver takes his foot off the gas pedal. It
takes 4.7 s for the car to slow down to 20 m/s.
How large is the net force slowing the car?
Answer in units of N.

Answer 1

By definition of average acceleration,

a = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ F = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.