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[tex]\huge\underline\bold\blue{ƛƝƧƜЄƦ}[/tex]

Given

[tex]\blue\star[/tex] v = 20m\s

[tex]\blue\star[/tex] a = 3m\s^2

[tex]\blue\star[/tex] t = 4sec

Firstly we have to find u

[tex]\star[/tex] a = [tex]\dfrac{v - u}{t}[/tex]

[tex]\star[/tex] 3m\s =[tex]\dfrac{20 - u}{4}[/tex]

[tex]\star[/tex]12m\s = 20 - u

[tex]\star[/tex]20 - u = 12m\s

[tex]\star[/tex]- u = -8

[tex]\star[/tex] u = 8

Now we can easily find distance by using second equation of motion

[tex]\red\star[/tex]s = ut + 1\2 at^2

[tex]\red\star[/tex]s = 8(4) + 1\2(3)(16)

[tex]\red\star[/tex]s = 32 + 24

[tex]\red\star[/tex]s = 56

So distance is 56 m\s hope it helps

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