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If $\left. p = -4.9t^2\ \middle|\ t \geq 0 \right.$, then $\frac{\partial p}{\partial t} = -9.8t$. How do I find $\frac{\partial p}{\partial t}$ in terms of $p$ rather than $t$ (i.e. velocity in terms of position)? Solving for $t$ in the first equation produces a square-root of a negative.

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No it doesn't, since $p$ is negative. –  André Nicolas Dec 15 '12 at 3:09
    
you're taking the positive root of t. it is p-- the negative one as André Nicolas says. –  ashley Dec 15 '12 at 3:22

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It is clear that $p\le 0$, so there is no square root of a negative here. We have $$t=\sqrt{\frac{-p}{4.9}}.$$ (Note we took the non-negative square root, since $t \ge 0$.)

Remark: Presumably you have dropped an object from some height, on an airless planet where the acceleration due to gravity is remarkably like Earth's.

You chose $p=0$ as the coordinate of the drop point. That means that the position at time $t\gt 0$ is negative. Perfectly fine. But if it interferes with the intuition, let $p=0$ at ground level.

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That presumption is quite correct. –  Hedgetrimmer Dec 15 '12 at 3:26

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