\begin{figure}[th] % From EMscript/figures_plt/minkowski_space_sol.
\centerline{\hbox{ \psfig{figure=smid1mink,width=10cm} }}
\caption{Minkowski space in which observer~B is at rest and
observer~A moves with speed $4c/5$ in negative $x$ direction. 
\label{fig_mid1mink} } \end{figure}

\item {\bf Travel in 2D Minkowski space.}

\begin{enumerate}

\item $\beta_1=60/75 = 4/5$.

\item The proper time of $B$ at the end of the first part of its travel 
is $\sqrt{(75)^2-(60)^2}=15\sqrt{5^2-4^2}=45$. This is also the time in 
the initial rest frame at which $B$ leaves the $x=0$ axis.

\item There are many ways to calculate the final meeting point. After
the first part of the travel at the initial rest frame $K'$ the position 
of $B$ is at $\pmatrix{45\cr 0}$. For the Lorentz transformation to this 
frame we use
$$ \gamma = \sqrt{\frac{1}{1-\beta^2}} = \sqrt{\frac{5^2}{5^2-4^2}}
   = \frac{5}{3}~~{\rm and}~~\beta\gamma = \frac{4}{3}\,.  $$
So, we find for the position of the final point in $K'$
$$ \pmatrix{x'^{\,0}\cr x'^{\,1}} =  \pmatrix{\gamma&-\beta\gamma 
   \cr -\beta\gamma & \gamma} \pmatrix{x^0\cr x^1} =  
   \pmatrix{5/3 & -4/3 \cr -4/3 & 5/3} \pmatrix{150\cr 0} =
   \pmatrix{250\cr - 200}\, . $$
See Fig.~\ref{fig_mid1mink}.

\item From the previous result we see that $B$ travels with the velocity
$$ \beta'\ =\ - \frac{200}{250}\ =\ -\frac{40}{41} $$
to catch up with $A$, $B$. The same result follows from the addition
theorem of velocities:
$$ -\beta'\ =\ -\frac{\beta+\beta}{1+(\beta)^2}\ =\
   -\frac{8}{5}\,\frac{25}{41}\ =\ -\frac{40}{41}\,. $$ 

\end{enumerate} \bigskip