# Proving the equations of motion

1. Jan 14, 2005

### misogynisticfeminist

If I got $$v=u+at$$

i get, $$\int v = \int (u+at) dt$$

which is $$s=ut+\frac {1}{2} at^2 + C$$

how do i explain away the integration constant in this derivation?

2. Jan 14, 2005

### dextercioby

You can't.It's x(0),the initial position.You can set it to zero,if u want by chosing the origin of the coordinate system in that very point.

Daniel.

3. Jan 14, 2005

### jdstokes

Why do you need to `explain away' the constant of integration. The constant $C$ corresponds to the position of the particle when $t = 0$. Thus you have
\begin{align*} x & = \frac{1}{2} a_xt^2 + u_xt + x_0 \\ x - x_0 & = \frac{1}{2} a_xt^2 + u_xt \\ s_x & = \frac{1}{2} a_x t^2 + u_xt. \end{align*}

4. Jan 14, 2005

### misogynisticfeminist

that's the explanation i needed lol, thanks alot......it didn't occur to me then.