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## Newton’s equations of motion – revisited

Last week, I showed how one could derive 3 of Newton’s equations of motion. As a colleague of mine pointed out to me on FaceBook, the 3rd equation I showed can, in fact, be derived using algebra from the 1st and 2nd. Also, if you go to the Wikipedia page on the equations of motion, you will find 5 listed, not just the 3 I showed. It turns out that the only two “fundamental” ones are the 1st and 2nd that I showed, the other three (including my 3rd, shown as the 4th in the image below) can be derived from the 1st and 2nd ones.

The 5 equations of motion as shown on the Wikipedia page

This is often the case in mathematics, there is more than one way to do something. So, although the method I showed to derive the 3rd equation is perfectly correct, and it also shows how we can rewrite $dv/dt \text{ as } (dv/ds \cdot ds/dt)$, which is a useful technique, I will today show how it can also be derived by algebraically combining equations (1) and (2).

Remember, equations (1) and (2) were

$v = u + at \text{ (Equ. 1)}$

$s = ut + \frac{1}{2} at^{2} \text{ (Equ. 2)}$

The first step is to square equation (1), which gives us

$v^{2} = (u + at)^{2} = u^{2} + 2uat + a^{2}t^{2} \text{ (Equ. 3a)}$

Next, we multiply equation (2) by 2 to get rid of the fraction

$2s = 2ut + at^{2} \text{ (Equ. 3b)}$

We next multiply each term in equation (3b) by $a$ to give

$2as = 2uat + a^{2}t^{2} \text{ (Equ. 3c)}$

Comparing equations (3a) and (3c) we can see that we can substitute the 2nd and 3rd terms of equation (3a) by $2as$ and so we have

$v^{2} = u^{2} + 2as \text{ (Equ. 3)}$

which is our equation (3) from the previous blog. As they say in mathematics, Quod erat domonstradum (QED) 🙂