# Need help in derivation

1. Oct 15, 2004

### Omar

Hello,

I'm new to the forums. This is my first thread here. I just wanted help in knowing where the motion equations (s=ut+0.5at^2, and v^2=u^2+2as) are derived from?

2. Oct 15, 2004

### Fredrik

Staff Emeritus
They are a consequence of the definition of acceleration as the second derivative of position (as a function of time). When the acceleration is constant, it's easy to find the velocity by doing a very simple integration.

$$x''(t)=a$$

$$x'(t)=at+v_0$$

The constant term is the velocity at time t=0. If we integrate this, we find the position as a function of time.

$$x(t)=\frac{1}{2}at^2+v_0t$$

I could have added a new constant term that would have represented the position at time t=0, but this is usually chosen to be zero.

The second equation in your post can be derived from the velocity and position equations above.

3. Oct 16, 2004

### Mahfuz

fundamentally they were derived from equation of acceleration.

:)

4. Oct 17, 2004

### ArmoSkater87

start off with the definition of acceleration...
$$a = \frac{v_f-v_0}{t}$$
$$v_f = v_0 + at$$

Definition of average velocity...
$$v_{ave} = d/t$$
$$d = \frac{v_f + v_0}{2}t$$

substitute...

$$d = \frac{(v_0+at) + v_0}{2}t$$
$$d = \frac{2v_0 + at}{2}t$$
$$d = v_0t + \frac{1}{2}at^2$$

For the second equation...it just a lot of manipulation of the equation above and a substitution of the definition of accleration.