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.