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I'm trying to build an intuition for the relationship of integrals to the equations of motion.

I was using this site to get an overall intuition for the relationship between integrals and derivatives. It seems clear how a cube changes in that the derivative is like adding 3 areas so that it grows by $3x^2$. And the reverse is just integrating the rate to get the cube $x^3 + c$.

Now my confusion is when we consider movement. Acceleration has units in $m/s^2$ but velocity is not cubed (at least in my understanding). So why does the integral of acceleration with respect to time produce values with cubes? I feel like I'm missing something obvious here.

adkane
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    Can you give a specific example where you get cubes? For example for constant acceleration $a$ the time integral is $at$ but no cube appears. – coffeemath Aug 09 '23 at 21:48
  • This might be my exact confusion, I'm taking acceleration to have units squared, and if this is integrated (power rule) I get units cubed. – adkane Aug 09 '23 at 21:53
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    It seems you think integrating means units get integrated. It is better to think in each case about units separately. For example to integrate $f(t)$ where $f(t)$ is in some units of "foo", one integrates just the function $f(t)$ w.r.t. $t$ and then think separately what the units should be by thinking about the units "foo" and trying to see what the integrated units should be. There isn't any definition of an integral of a random unit that I know of. – coffeemath Aug 09 '23 at 22:01
  • That is helpful. Can you see where my confusion stems from? If I'm dealing with a cube like in the link, I'm in cubic space and differentiating that gets me to squared space. – adkane Aug 09 '23 at 22:06
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    The answer of CW279 seems to cover things. – coffeemath Aug 09 '23 at 22:17
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    FWIW, the power of $\mathrm{s}$ does increase from $-2$ to $-1$. – L. F. Aug 10 '23 at 09:01

1 Answers1

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If you integrate a function $f(t)$ with respect to time $t$, the result will be measured in the units of $f$, multiplied by the units of $t$; ie.

$ \begin{equation} [\int f(t)\,dt] = [f(t)]\times [t] \end{equation} $

where $[\cdot]$ means "the units of". Likewise, if you differentiate $f(t)$ with respect to $t$, the result will have the units of $f(t)$, divided by the units of $t$:

$ \begin{equation} [\frac{df}{dt}] = \frac{[f(t)]}{[t]} \end{equation} $

As you noted, in SI units acceleration has the units of $\mbox{m/s}^2$. So if you integrate $a(t)$ to find the velocity $v(t)$, you know the units of $v(t)$ will be $\mbox{m/s}$:

$ \begin{equation} [v(t)]=[\int a(t)\,dt] = [a(t)]\times [t] = \mbox{m/s}^2\times s=\mbox{m/s} \end{equation} $

However, this does not imply the acceleration must scale as some particular power $n$ of $t$, since any power of $t$ can be multiplied by some dimensional prefactor $A$ such that $At^n$ has units of $\mbox{m/s}^2$. For example, if $n=3$, and the units of $A$ are $\mbox{m/s}^5$, then $At^3$ is a valid acceleration, in the sense that its units are $\mbox{m/s}^2$.

CW279
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  • This is a great response. Why would t have different powers? Is this just a fact of the scale that you're working with? – adkane Aug 09 '23 at 22:18
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    Glad to help! If you think of Newton's second law, then $a(t)=F(t)/m$, where $m$ is the mass of your system and $F(t)$ is the force acting on it. Since $m$ doesn't depend on $t$ (in this case), then different powers of $a$ correspond to different powers of $F$; ie. different types of force acting on your system. If $n=0$ then the force doesn't depend on time - this might correspond to free-fall in Earth's gravitational field, where $F=mg$. If $n=1$ then the force increases linearly in time, and so forth. – CW279 Aug 09 '23 at 22:24
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    @CW279 In your comment above what is $n$ in "If $n=0$ then..."? [n not explained in the comment...] – coffeemath Aug 09 '23 at 22:48
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    @coffeemath Good catch- $n$ is the power to which $t$ is raised; $a(t)=At^n$. I included it in the body of my answer, but I forgot to put it in my comment. – CW279 Aug 09 '23 at 23:25