![\huge\underline\bold\blue{ƛƝƧƜЄƦ}](https://img.qammunity.org/2021/formulas/mathematics/high-school/5vp1i0ypqpff2rkpt36o87cpm7rffbhqco.png)
Given
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a =
![(v - u)/(t)](https://img.qammunity.org/2021/formulas/physics/high-school/dvu2uw9kw1928nl514dn84m1iq81zjvkp9.png)
3m\s =
![(20 - u)/(4)](https://img.qammunity.org/2021/formulas/physics/high-school/y23n772la9i8fzm466t7mpbthvpe3p937t.png)
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps