# Derivative of position is velocity and of velocity is acceleration?

This is more of a personal question (i.e. not school related), but how has it been proven that the derivative of position is velocity and derivative of velocity is acceleration? I did some Google searching on this and everyone just states it as fact without any proof behind it.

So, does anyone know of a resource that goes about proving this fact?

Thanks!

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It's a definition, not a proof. –  Mario Carneiro Dec 16 '12 at 17:47
There's no such thing as definition without a proof in mathematics. Everything can be proved somehow and I'm also looking for an analitic answer to the question, the conceptial one (given by @Jasper Loy) can be reached by drawing a graph and letting the brain figure it out. –  Marco Aurélio Deleu Mar 17 '13 at 14:08
@MarcoAurélioDeleu That's complete nonsense. Proofs are needed to justify logical statements, but definitions aren't logical statements. Definitions aren't true or false or provable or unprovable. Definitions allow us to make shorthands for bundles of hypotheses (for instance, the definition of a circle in geometry.) You can't prove the definition of a circle. –  rschwieb Sep 30 '13 at 20:15
It does not need proof. I could say "let the amount of tinned food you buy per week be called Tintin!" That does not need a proof of any sort does it. It is just a definition / statement to allow us to use it in speech, calculations, etc. In terms of derivatives, after velocity there is acceleration, jerk, jounce, crackle, pop,... The derivative list is endless. –  pnizzle Jul 4 at 5:52

Velocity is defined as the rate of change of displacement with respect to time, and acceleration is defined as the rate of change of velocity with respect to time.

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If we define change in position as $\delta x$ and a change in time as $\delta t$, then we, know that the average velocity over that time is given by $$\frac{\delta x}{\delta t}.$$ But, we do not want to know the average velocity, so we want to find the instantaneous velocity. In this case, we, by the definition of the derivative, want to take the limit as $\delta t\rightarrow0$. So, we have as velocity $$\lim_{\delta t \to 0} \frac{\delta x}{\delta t}=x'(t)$$ as desired. This isn't really a proof, per se, but it is a deeper explanation which perhaps you are seeking. A similar explanation applies to acceleration.

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The derivative is the slope of the function. So if the function is $f(x)=5x-3$, then $f'(x)=5$, because the derivative is the slope of the function. Velocity is the change in position, so it's the slope of the position. Acceleration is the change in velocity, so it is the change in velocity. Since derivatives are about slope, that is how the derivative of position is velocity, and the derivative of velocity is acceleration. So if the position can be expressed with the function $f(x)=x^2 - 3x + 7$, then the derivative would be $f'(x)=2x-3$ since that is the slope of the function at any given point, and since it is the slope of the position function, it is velocity. Same for acceleration; $f"(x)=2$, which is the derivative of velocity, which makes it slope. The slope of velocity is acceleration. This is how the derivative of position is velocity and the derivative is position.

NOTE: These functions are entirely hypothetical and were created on the spur of the moment.

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A great place to start would be to purchase any college level physics or calculus textbook. You are not simply going to find the answer on google, however, the derivative function proof is basically first day of college physics and calculus eventually gets around to the proof. Hope this helps.

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