
Given
 v = 20m\s
 v = 20m\s
 a = 3m\s^2
 a = 3m\s^2
 t = 4sec
 t = 4sec
Firstly we have to find u
 a =
 a = 

 3m\s =
 3m\s =

 12m\s = 20 - u
12m\s = 20 - u
 20 - u = 12m\s
20 - u = 12m\s
 - u = -8
- u = -8
 u = 8
 u = 8
Now we can easily find distance by using second equation of motion
 s = ut + 1\2 at^2
s = ut + 1\2 at^2
 s = 8(4) + 1\2(3)(16)
s = 8(4) + 1\2(3)(16)
 s = 32 + 24
s = 32 + 24
 s = 56
s = 56
So distance is 56 m\s hope it helps