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QUESTION

If a particle $P$ has position \textbf{r} and its acceleration at
time $t$ is given by $\ds\frac{d^2\mathbf{r}}{dt^2}=-g\mathbf{k}$,
where $g$ is a constant, find the general form for the particle's
position at time $t$.

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ANSWER

$\ds\frac{d^2\mathbf{r}}{dt^2}=-g\mathbf{k},\
\frac{d\mathbf{r}}{dt}=-gt\mathbf{k}+\mathbf{c}_1,\
\mathbf{r}=-\frac{gt^2}{2}\mathbf{k}+t\mathbf{c}_1+\mathbf{c}_2,$
where $\mathbf{c}_1$ and $\mathbf{c}_2$ are constant vectors.




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