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A few days ago, I realized a similarity between distance with constant acceleration, $d = v_i t + 1/2 a t^2$, and the sum of integers up to n, $(n^2 + n)/2$. This came up again today when I decided to work out some formulas for distance and velocity with constant acceleration updated at discrete intervals, as happens in physics simulations I've programmed.

Discretely incremented $v$ and $x$ with constant acceleration, $v_i = 0$

  • Add $a\mathrm{d}t$ to $v$ every tick
  • Add $v\mathrm{d}t$ to $x$ every tick

$$\begin{align} v_f &= a\mathrm{d}t + a\mathrm{d}t + a\mathrm{d}t + \cdots \\ x_f &= v_1\mathrm{d}t + v_2\mathrm{d}t + v_3\mathrm{d}t + \cdots \\ v_2 &= (a\mathrm{d}t) + (a\mathrm{d}t) = 2(a\mathrm{d}t) \\ v_n &= n(a\mathrm{d}t) \\ x_f &= (a\mathrm{d}t)\mathrm{d}t + 2(a\mathrm{d}t)\mathrm{d}t + 3(a\mathrm{d}t)\mathrm{d}t + \cdots \\ x_n &= \sum_{i=0}^{n}i(a\mathrm{d}t)\mathrm{d}t = \frac{n^2 + n}{2}(a\mathrm{d}t)\mathrm{d}t \\ n &= \left\lfloor\frac{t_\text{total}}{\mathrm{d}t}\right\rfloor \\ x_f &= \frac{1}{2}\Biggl[\biggl(\frac{t_\text{total}}{\mathrm{d}t}\biggr)^2 + \frac{t_\text{total}}{\mathrm{d}t}\Biggr](a\mathrm{d}t^2) = \biggl(\frac{t_\text{total}^2}{2\mathrm{d}t^2} + \frac{t_\text{total}}{2\mathrm{d}t}\biggr)(a\mathrm{d}t^2) \\ x_f &= a\mathrm{d}t^2\times\frac{1}{2}\times\frac{t_\text{total}^2}{\mathrm{d}t^2} + \frac{1}{2}a\mathrm{d}t\frac{t_\text{total}}{\mathrm{d}t} \\ x_f &= \frac{1}{2}at_\text{total}^2 + a\mathrm{d}t\frac{t_\text{total}}{2} \\ \lim_{\mathrm{d}t\to 0} &= \frac{1}{2}at^2 = \text{normal $v_f$ equation where $v_i = 0$ and $\Delta a = 0$} \end{align}$$

I'd like to know what relations to theory this or an idea like it has if any, and if it touches on anything else.

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@Kurt: Are you aware of integral calculus? – Qmechanic May 9 '13 at 22:36
up vote 17 down vote accepted

The result you've got would be better known as this:

$$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$

In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn.

The only difference is that you've done it explicitly, using limits, rather than using the rules for integrating polynomials. That's a good thing! It will help you understand where the formula comes from and what it means, and if you continue to do more with numerical integration (as in your simulations), it's going to be very useful to know the details of how this stuff works.

Now, considering that this has been known for about 350 years, its applications have been pretty thoroughly explored. It's a part of classical kinematics, which is a branch of physics that analyzes simple motion without any quantum effects, so there is no special significance to the Planck time with respect to this equation.

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Thanks for the answer and formatting! The only physics I've done in school didn't have any calculus, but I've learned some of it for myself. I wasn't really aware integrals could be done with limits as derivatives can, though, which is good to know. The material I'd read about calculus from always took it from the angle of doing the opposite of a derivative, just applying the derivative rules backwards. – Kurt May 8 '13 at 10:56
@Kurt: There are lots of ways of defining integrals, but the most straightforward one you'd want is the Riemann integral. This kind of formalization using limits is only ~150 years old, despite the result well-being known for a lot longer as David says. (The limit definition of derivative was by Cauchy, and is likewise younger than calculus.) – Stan Liou May 9 '13 at 22:30
@Kurt: Integrals are often used as the inverse operation of derivatives, which is probably why your calculus reading material introduces it that way - but they're also defined using limits, as in the article Stan linked to. I would definitely echo the recommendation to read that, or read some other resource which defines the Riemann integral as a limit, because for physics it's important to know both interpretations and be able to switch between them. (The fact that the Riemann integral undoes a derivative is known as the first fundamental theorem of calculus.) – David Z May 10 '13 at 0:15
I actually do remember reading about the definition of an integral as the Riemann sums earlier, I just didn't consciously remember it much the day I typed that response for some reason. Since then I've been revisiting it and it's pretty intuitive. I just used some integrals to derive the volume of a sphere and hypersphere and other dimensional spheres, integrals feel natural to me. I'll be sure to read more material. – Kurt May 10 '13 at 0:54

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