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# A particle is moving with the given data. Find the position of the particle.$a(t) = 2t + 1$, $\quad s(0) = 3$, $\quad v(0) = -2$

## $$s(t)=\frac{1}{3} t^{3}+\frac{1}{2} t^{2}-2 t+3$$

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So we're giving a particle with the following data acceleration that's a F. T equals two T plus one. So when we find its velocity velocity is going to be the anti derivative of acceleration. So it's going to be um t squared plus T. And then we're gonna have plus a constant. But we know that via zero is negative two. That means this is gonna be minus two. Our position function at safety, it's going to be able to t cube over three. Okay. Plus t squared over two and then we're going to have minus two T. And um plus a constant. But we know that s of zero is three. So that means our constant value is three. Therefore this is our final position function.

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