Proving the equations of motion

[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?

 

[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.

 

[jdstokes]

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?

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
$\begin{align*}<br /> x & = \frac{1}{2} a_xt^2 + u_xt + x_0 \\<br /> x - x_0 & = \frac{1}{2} a_xt^2 + u_xt \\<br /> s_x & = \frac{1}{2} a_x t^2 + u_xt.<br /> \end{align*}$

 

[misogynisticfeminist]

Why do you need to `explain away' the constant of integration. The constant $C$ corresponds to the position of the particle when $t = 0$.

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