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I am thinking about the links between SUVAT equations (constant acceleration), and equations for motion when higher-order measurements are constant (for example, when jerk is constant, or snap is constant). See here for an excellent discussion of how to derive the jerk equations using calculus.

I am attempting to teach this topic first without Calculus, and then to revisit it later once the students are aware of differentiation and integration. They understand informally that calculating the gradient of velocity function gives you acceleration, and finding the area underneath undoes this process, but have no specific knowledge about how to differentiate or integrate.

For example, we can derive the SUVAT equation below by calculating the area of the relevant trapezium on the velocity-time graph (see below)

$$s = ut + \frac{1}{2}at^2$$

Is there any way that derives the third jerk equation, without using formal calculus?

$$s = ut + \frac{1}{2}a_u t^2 + \frac{1}{6}jt^3$$

In this equation, $s$ is displacement, $u$ is initial velocity, $t$ is time, $a$ is initial acceleration, and $j$ is constant jerk.

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I've never seen the jerk discussed in any non-calculus physics course. An easy way to do it without using geometry is the following.

First, the variation of $t^2$ is $\Delta t^2=(t+h)^2-t^2\approx 2th$ and $\Delta t^3\approx 3t^2h$ for small $h$. Second, the velocity is to the displacement what the acceleration is to the velocity and what the jerk is to the acceleration. Third, in your last equation $\frac{\Delta s}{\Delta t}=u+at+\frac12 jt^2$ which is identical to your first equation.

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  • $\begingroup$ I understand that it is a bit weird to do it this way around. The students are 17 year olds, and just so happen to learn Mechanics before Calculus. We are trying to think of ways to weave in informal ideas about gradient/area as early as possible. Thank you for your reply! Ideally it would use geometry. I like your idea of using small changes of t - nice to not have to rely on any knowledge of how to differentiate $\endgroup$ – Jezza Judge Dec 1 '17 at 16:09
  • $\begingroup$ @JezzaJudge I think you can do it graphically without integrals. You can find the area under a parabola using (Riemann) sums. You only need to prove first that $\sum\limits_{i=1}^{n}i^2=\frac16 n(n+1)(2n+1)$. $\endgroup$ – Paracosmiste Dec 2 '17 at 14:31
  • $\begingroup$ Nice! Thank you. $\endgroup$ – Jezza Judge Dec 4 '17 at 6:51

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