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

 

[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 alot...it didn't occur to me then.