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I understand that acceleration as a function of time is the derivative of velocity with respect to time, but how can that still be so if acceleration is a function of position or velocity?

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Acceleration is the derivative of velocity with respect to time, by definition.

It doesn't matter what factors affect it, that's still what it is. You could have acceleration as a function of the amount a spring is stretched, acceleration as a function of how much you press the gas pedal, acceleration of a sail boat as a function of how fast the wind is blowing, or whatever, and acceleration would still be defined as the derivative of velocity with respect to time.

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If you are able to specify acceleration as a function of position or velocity, then it is still true that $$a=\frac{\text dv}{\text dt}=\frac{\text d^2x}{\text dt^2}$$ because this is just the definition of acceleration.


You encounter the position case in things like simple harmonic motion (like a mass on a spring). Then you can think of acceleration as a function of position where $$a(x)=-\frac kmx$$ Solving the differential equation gives us what $x(t)$ is $$x(t)=A\cos(\omega t+\phi)$$ where $\omega^2=k/m$, and $A$ and $\phi$ depend on initial conditions

You encounter the velocity case when considering drag forces. If a drag force is present that is proportional to the velocity, then we have $$a(v)=-\frac bmv$$ Solving the differential equation gives us what $v(t)$ is $$v(t)=v_0\,e^{-b/m\,t}$$ where $v_0$ depends on the initial conditions.

You can also have both cases at the same time. For example, the damped harmonic oscillator: $$a=-\frac kmx-\frac bmv$$ or really in many other scenarios. In general you could have a differential equation of the form $$a(x,v)=f(x)+g(v)$$ which you could write as a differential equation to solve for $x(t)$ as $$\ddot x=f(x)+g(\dot x)$$


You can also have neither case. If your system is exhibiting some weird motion (like if I am just randomly moving a box back and forth across the ground), then there isn't even single-valued function $a(x)$ or $a(v)$ for this motion. However, there will always be $a(t)$.

Ultimately the time derivatives will always be the definition of acceleration, but this definition of acceleration does not depend on what we decide to express acceleration as a function of.

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  • $\begingroup$ Thanks for the input guys! Thinking about this intuitively though, say our function for velocity is 3t^2 for simplicity. If we take the derivative of velocity with respect to time, we will obtain 6t. This is the only answer. Now by the logic that acceleration with respect to time, position or whatever else would both be equivalent functions equal to 6t. How does this make sense? (my logic seems flawed already as I type it out, but I'm not sure how to put the incorrectness into words). $\endgroup$ – Joey Chen Aug 27 at 3:13
  • $\begingroup$ @JoeyChen If $v(t)=3t^2$ and $a(t)=6t$, then in terms of time and velocity $a=3vt$ I suppose. Or if $x(0)=0$ then $x(t)=t^3$ and so $a=18x/v$ (might have a divide by $0$ issue for $t=0$, but oh well). But how we choose to mathematically represent our functions doesn't change anything about them physically. $\endgroup$ – Aaron Stevens Aug 27 at 3:19
  • $\begingroup$ @JoeyChen Also, remember to up vote any useful answers and to eventually select an answer as the accepted answer. $\endgroup$ – Aaron Stevens Aug 27 at 3:23

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