The given equation is $\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}}$. We can rewrite $\frac{2}{t^{3}}$ as $2t^{-3}$. Now, we integrate $2t^{-3}$ with respect to $t$ to find the first derivative of $r$ with respect to $t$, $r'(t)$:
$$\int 2t^{-3} dt = -t^{-2} +
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