= Solution
Differentiate each segment and express the <four-velocity> using a continuous <rapidity>:
$$
U(s)=c(\cosh\theta(s),\sinh\theta(s),0,0),\qquad
\theta(s)=\begin{cases}
\omega s,&0\leq s<T,\\
\omega(2T-s),&T\leq s<3T,\\
\omega(s-4T),&3T\leq s\leq4T.
\end{cases}
$$
At $T$ both rapidities are $\omega T$; at $3T$ both are $-\omega T$. Hence \b[the <four-velocity> is continuous at both junctions]. This expression also gives $g(U,U)=c^2$ everywhere, so the parameter remains <proper time> on the later segments.
Away from the junctions,
$$
A(s)=c\theta'(s)(\sinh\theta(s),\cosh\theta(s),0,0),
$$
with $\theta'$ successively $+\omega,-\omega,+\omega$. The rapidity is continuous but its slope reverses. Thus \b[the <four-acceleration> has a finite jump to its negative at each junction], while its magnitude stays $c\omega$ on either side. There is no velocity jump; the idealized path is continuously differentiable but not twice differentiable at the switching events.
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