A 2380 kg car traveling to the west at 20.1 m/s slows down uniformly under a force of 8820 N to the east.

a) How much force would be required to cause the same acceleration on a car of mass 3300 kgb) how far would the car move before stopping?

a) Force would be required to cause the same acceleration on a car of mass 3300 kg is 12229.41 N to the east. b) Distance traveled before stopping is 54.45 m

Explanation:

a) Force = 8820 N

Mass = 2380 kg

We have

Force = Mass x Acceleration

-8820 = 2380 x a

a = - 3.71 m/s²

Now the mass is 3300 kg

Force = Mass x Acceleration

F = 3300 x - 3.71 = - 12229.41 N

Force would be required to cause the same acceleration on a car of mass 3300 kg is 12229.41 N to the east.

b) We have equation of motion v² = u² + 2as

Initial velocity, u = 20.1 m/s

Acceleration, a = - 3.71 m/s²

Final velocity, v = 0 m/s

Substituting

v² = u² + 2as

0² = 20.1² + 2 x - 3.71 x s

s = 54.45 m

Distance traveled before stopping is 54.45 m