Why is acceleration as a function of position or velocity equal to the derivative of velocity with respect to time?

Question

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?

Answer 1

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.

Answer 2

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.

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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 $$\ddot x=f(x)+g(\dot x)$$

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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.