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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.

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) đź™‚

### 4 Responses

1. very. nice gooood.

2. Very neat explanation

3. […] for a variety of angles in between. You can also check the theoretical best launch angle using the equations of motion in physics, a perfect activity for older […]

4. on 03/07/2020 at 14:27 | Reply Naveed Raza

Good Experience gained by me after visiting this website.