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## Derivation of Newton’s equations of motion

Anyone who has studied mechanics / dynamics will have come across Newton’s equations of motion (not to be confused with his laws of motion). The ones I get my students to use are

1. $v = u + at \; \; \text{ (Equ. 1)}$
2. $s = ut + \frac{1}{2} at^{2} \; \; \text{ (Equ. 2)}$
3. $v^{2} = u^{2} + 2 as \; \; \text{ (Equ. 3)}$

where $u$ is the initial velocity, $v$ is the velocity at time $t$, $s$ is the displacement and $a$ is the acceleration. Note, these equations are only true for constant acceleration, but that actually covers quite a lot of situations. They can all be derived from the definition of acceleration.

Newton’s equations of motion can be derived from his 2nd law of motion.

## Derivation of Equation 1

We start off with out definition of acceleration, which is the rate of change of velocity. Writing that mathematically,

$a = \frac{dv}{dt}$

This is an example of a first order differential equation. To solve it we integrate. So we have

$a \int dt = \int dv$

When we integrate without limits, we have to include a constant term, so we can write

$at = v + C$

where $C$ is our constant. To determine the value of the constant we need to put in some conditions, such as (but not necessarily) initial conditions. When $t=0$ we have defined that $v=u$, so we can write

$0 = u + C \rightarrow C=-u \rightarrow \boxed{v = u + at \; \; \text{ (Equ. 1)} }$

## Derivation of Equation 2

To derive equation two, which we notice involves the displacement (the vector equivalent of distance), we do the following

$v = \frac{ds}{dt} = u + at \rightarrow ds = \int u dt + \int at dt$
$s = ut + \frac{at^{2}}{2} + C$

When $t=0 \text{ then } s=0$ so we can write

$0 = 0 + 0 +C \rightarrow C=0 \rightarrow \boxed{s = ut + \frac{1}{2}at^{2} \; \; \text{ (Equ. 2)} }$

## Derivation of Equation 3

To derive equation 3 we use the trick of writing the acceleration $a$ in terms of the velocity $v$ and the displacement $s$. To do this we write

$a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v = v \frac{dv}{ds}$

So, writing

$v \frac{dv}{ds} = a \rightarrow \int v dv = \int a ds \rightarrow \frac{v^{2}}{2} = as + C$
Again, we can work out $C$ by remembering that $s=0 \text{ when } t=0$ so

$\frac{u^{2}}{2} = 0 + C \rightarrow C = \frac{u^{2}}{2}$

and so

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

### 13 Responses

1. […] 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 […]

2. […] this equation also allows us to derive the three equations of motion, equations like and , as I did in this blog. It tells us that it is more difficult to accelerate a more massive object than it is a less […]

3. […] just comes from Newton’s 2nd equation of motion , see my blog here which derives those […]

4. Thise is a very most impartant of studen.

5. Simple and catchy,is ur lectures. Thnk u

6. Thank you. This was very useful to me

7. Thank you it was very helpful for me.

8. It’s surprising to find on wordpress.com a resource so precious about equations.

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9. Thanks

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11. Tqu u it was very useful to me……

12. How many ways to find out these equation

13. Pls can you explain more further how you got equation 2