# Relationship between acceleration, velocity and position

I'm learning some applications for equation of motion. But I'm failing to relate velocity, acceleration and position.

If $v=\frac{dr}{dt}$ and $a=\frac{dv}{dt}$, why $a$ is $\frac{d^2r}{dt^2}$ instead of $\frac{dr}{dt^2}$? Probably i'm lacking basic calculus or physics.

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"Probably i'm lacking basic calculus" Yep. Now would be a good time to review the topic. Or if you haven't had it yet, to seek out a "algebra based" introduction to physics in the mean time. There are some very clear explanations of the basic kinematic equations that do not require calculus, and you'll likely be less confused if you try them. Otherwise you'll flounder on stuff that the author thought was clear because (s)he made some assumptions about your mathematical preparation. – dmckee May 13 '12 at 21:25

## 1 Answer

The answer is in front of you

$$a = \frac{{\rm d}(u)}{{\rm d} t} = \frac{{\rm d}}{{\rm d} t} \left( \frac{{\rm d}(r)}{{\rm d} t} \right) = \frac{{\rm d}^2(r)}{{\rm d} t^2}$$

as a matter of convention. It is just the way we write 2nd derivatives.

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You might even go so far as to format the basic differentiation in time operator as $\frac{\mathrm{d}}{\mathrm{d}t}(u)$. – dmckee May 13 '12 at 21:26