Answer:
(a) Time will be t = 3.56 sec
(b) Distance traveled by car when they are side by side is 37.38712 m
(b) Velocity of race car = 21.004 m/sec
velocity of stock car = 12.816 m/sec
Explanation:
We have given acceleration of the car 
Acceleration of the stock car 
When 1st car overtakes the second car then distance traveled by both the car will be same
(a) So 
As both car starts from rest so initial velocity of both car will be 0 m/sec
It is given that stock car leaves 1 sec before
So 
After solving t = 3.56 sec
(b) From second equation of motion 
(c) From first equation pf motion v = u+at
So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec
Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec