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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf SOLUTIONS -- SET 12}\\%
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\noindent
Problem 45:
\bigskip

\noindent
In units with $c=1$ and time, distance in years:
Always $t_{\rm news}=t-x$.
\bigskip

\noindent
In the first 5 years (time unit years, distance unit light years):
\begin{eqnarray} \nonumber
  \zeta &amp;=&amp; \alpha\,\tau\,,~~\alpha=g/c\,,~~0\le\tau\le 5\,y~~
  ({\rm accelerating\ away\ from\ earth}\,,\\
  t &amp;=&amp; \sinh(\zeta)/\alpha\,, \nonumber \\ \nonumber
  x &amp;=&amp; \left[\cosh(\zeta)-1\right]/\alpha\,.\\ \nonumber
\end{eqnarray}
With $\tau_1=5\,y$, $\zeta_1=\alpha\,\tau_1$, $t_1=t(\zeta_1)$ and 
$x_1=x(\zeta_1)$ we have in the next five years
\begin{eqnarray} \nonumber
  \zeta &amp;=&amp; \zeta_1-\alpha\,(\tau-\tau_1)\,,~~5\le\tau\le 10\,y
  ~~({\rm slowing\ down})\,,\\
  t &amp;=&amp; t_1+[t_1-\sinh(\zeta)/\alpha]\,, \nonumber \\ \nonumber
  x &amp;=&amp; x_1+\{x_1-\left[\cosh(\zeta)-1\right]/\alpha\}\,.
\end{eqnarray}
With $\tau_2=2\,\tau_1$, $t_2=2\,t_1$ and $x_2=2\,x_1$ we have in the 
next five years
\begin{eqnarray} \nonumber
  \zeta &amp;=&amp; \alpha\,(\tau_2-\tau)\,,~~10\le\tau\le 15\,y
  ~~({\rm accelerating\ towards\ earth})\,,\\
  t &amp;=&amp; t_2-\sinh(\zeta)/\alpha\,, \nonumber \\ \nonumber
  x &amp;=&amp; x_2-\left[\cosh(\zeta)-1\right]/\alpha\,.
\end{eqnarray}
With $\tau_3=3\,\tau_1$, and $t_3=3\,t_1$ we have in the final five 
years ($\zeta_3=\zeta_1,\ x_3=x_1$)
\begin{eqnarray} \nonumber
  \zeta &amp;=&amp; \zeta_1+\alpha\,(\tau-\tau_3)\,,~~15\le\tau\le 20\,y
  ~~({\rm slowing\ down})\,,\\
  t &amp;=&amp; t_3+t_1+\sinh(\zeta)/\alpha\,, \nonumber \\ \nonumber
  x &amp;=&amp; x_1-\left[\cosh(\zeta)-1\right]/\alpha\,.
\end{eqnarray}
Remark: News from the spaceship, sent of at speed of light at its
time $\tau$, reaches earth at time~~$t_{\rm\ to\ earth}=t+x$.

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% \caption{Ladder for assignment~14.} \label{fig_lader} \end{figure}

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