Ok thank you all I can confirm Keithth G's solution.
Without this solution I felt to unsure to solve it but now I got it too (hopefully):
I
[math]
\begin{split}
a&=j*t\\
v&=\frac{ds}{dt}\\
a&=\frac{dv}{dt}\\
j&=\frac{da}{dt}=const(j)\\
\end{split}
[/math]
II
[math]
\begin{split}
s&=\int v\,dt\\
v&=\int a\,dt\\
a&=\int j\,dt=j\,t\,\,\, (for\,j=const)\\
\end{split}
[/math]
III
Because:
[math]
\begin{split}
\int t^n\,dt=\frac{1}{n+1}\,t^{n+1}
\end{split}
[/math]
We can say...
[math]
\begin{split}
v&=\int j\,t\,dt=j\int t\,dt\\
&=\frac{j}{2}\,t^2\,\,\, (for\,V_{0}=0)\\
\end{split}
[/math]
so let's insert this in the formular for s of II
IV
[math]
\begin{split}
s&=\frac{j}{2}\int t^2\,dt\\
&=\frac{j}{2} \,\frac{1}{1+2}\,t^{2+1}\\
&=\frac{1}{6}\,j\,t^3
\end{split}
[/math]
Now using
[math]
\begin{split}
v&=\frac{j}{2}\,t^2\,\,\, \\
t&=\sqrt{\frac{2\,v}{j}}
\end{split}
[/math]
so...
[math]
\begin{split}
t^3&=(\sqrt{\frac{2\,v}{j}})^3\\
&=\sqrt{\frac{2^3\,v^3}{j^3}}\\
&=\sqrt{\frac{8\,v^3}{j^3}}\\
\end{split}
[/math]
insert this for t^3 this into last line of IV:
[math]
\begin{split}
s&=\frac{1}{6}\,j\,\sqrt{\frac{8\,v^3}{j^3}}\\
\sqrt{\frac{8\,v^3}{j^3}}&=\frac{6\,s}{j}\,\,\,\,|^2\\
\frac{8\,v^3}{j^3}&=36 \frac{s^2}{j^2}\,\,\,\,|*\frac{j^3}{8}\\
\end{split}
[/math]
now we want to know v^3
[math]
\begin{split}
v^3&=\frac{36\,s^2\,j^3}{8\,j^2}\\
&=\frac{9}{2}s^2\,j \,\,\,\,|\sqrt[3]{} \\
\end{split}
[/math]
...
[math]
\begin{split}
v&=\sqrt[3]{\frac{9}{2}\,s^2\,j} \,\,\,\,|*\frac{4}{4} \\
&=\frac{\sqrt[3]{36}}{2}\,\sqrt[3]{s^2j}\\
&=1,650...\,\sqrt[3]{s^2j}
\end{split}
[/math]
:tiphat:
