a) For the motion of car with uniform velocity we have ,
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting

The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting

Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209 
38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.
A or possibly C because the other options have nothing to do with the size of the vibration. If i was you I would answer with A
Explanation:
If force or distance is increased, then amount of workdone will also increase.