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Find the position of the particle. a(t) = 2t + 3, s(0) = 8, v(0) = −3 A  particle is moving with the given data. Find the position of the particle. a(t) = 2t + 3, s(0) = 8, v(0) = −3

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`(d(v(t)))/(dt) = a(t)` , `(d(s(t)))/(dt) = v(t)`

So v(t) is the antiderivative of a(t) and s(t) is the antiderivative of v(t).

`v(t) = t^2 + 3t + C` , since `v(0) = -3 = 0^2 + 3(0) + C` , so `C = -3`

`v(t) = t^2 + 3t -...

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`(d(v(t)))/(dt) = a(t)` , `(d(s(t)))/(dt) = v(t)`

So v(t) is the antiderivative of a(t) and s(t) is the antiderivative of v(t).

`v(t) = t^2 + 3t + C` , since `v(0) = -3 = 0^2 + 3(0) + C` , so `C = -3`

`v(t) = t^2 + 3t - 3`

`s(t) = t^3/3 + 3/2t^2 - 3t + C` , since `s(0) = 8 = 0^3/3 + 3/2(0)^2 - 3(0) + C` , `C = 8`

so my final answer is

`s(t) = t^3/3 + 3/2t^2 - 3t + 8`

 

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