\documentclass[12pt]{article} % BB Feb 22 2009.
\usepackage{epsfig}
\textheight 9.0in \textwidth  6.0in \topmargin -0.4in \headsep 0.4in  
\oddsidemargin  0.25in  \evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
   \vrule width.#2pt} \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
\pagestyle{empty}

\begin{document}

\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf SOLUTIONS -- SET 11}\\%
\end{center}
\setcounter{equation}{0}

\begin{description}

\item{(39)} For constant $\alpha$:
$$ t_2-t_1 = \int_{\tau_1}^{\tau_2} \cosh(\alpha\tau)\,d\tau =
   \alpha^{-1}\left[\sinh(\alpha\tau_2)-\sinh(\alpha\tau_1)\right]\,.$$
With $\alpha^{-1}=c/g=3\times 10^8/9.81\,[s]$, $\tau_2=5\,[y]$, $\tau_1
=0\,[y]$, $1\,[y]=365\times 24\times 3600\,[s]$ one finds $t_2-t_1=
84.12\,[y]$. The same for the other three sections of the trip. 
Therefore, the year on earth is $2020+4\times 84.12=2020+336.47 \to 2356$.
\end{description}

% \begin{figure}[tb] \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=figs/ladder.eps,width=7.5cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader} \end{figure}

\end{document}
