User:Asjafs/sandbox25

3D Cubic Positional Velocity System

In the system you determine a cubic area and divide it in to 1x1x1cm blocks. You then determine the x, y and z coordinates of the object. For example if at $t_{\mathrm {0} }$ the object is at $(0,0,0)$, at $t_{\mathrm {1} }$ the object is at $(1,1,1)$ and at $t_{\mathrm {2} }$ the object is at $(2,2,2)$. First you calculate the displacement between $(0,0,0)$ and $(1,1,1)$:

$d_{\mathrm {01} }={\sqrt {(1-0)^{2}+(1-0)^{2}+(1-0)^{2}}}={\sqrt {1+1+1}}={\sqrt {3}}=1.732cm$

Then you calculate the displacement between $(1,1,1)$ and $(2,2,2)$:

$d_{\mathrm {12} }={\sqrt {(2-1)^{2}+(2-1)^{2}+(2-1)^{2}}}={\sqrt {1+1+1}}={\sqrt {3}}=1.732cm$

Then you calculate the total displacement:

$d_{\mathrm {total} }=d_{\mathrm {01} }+d_{\mathrm {12} }={\sqrt {3}}+{\sqrt {3}}=3.464cm$

Then since this happens over $2s$ you calculate:

${\frac {3.464cm}{2s}}=1.732cm/s$

This means that the object is moving at $1.732cm/s$.