A2380 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?

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

the area between the graph and the t axis