Special relativity: can the velocity of a free object be discontinuous relative to an accelerated observer who stops accelerating? Suppose an observer   ${\it {\mathcal O}}$     leaves point O*, origin of  the Minkowski frame K*,  in the $O^{*}x^{*} $ direction, with a uniform acceleration $a_{0} $ (as measured by himself) towards an object ${\it {\mathcal P}}$, fixed with regard to O*,   located at $x_{0}^{*} $ on that axis. In K*, the ``hyperbolic motion'' of ${\it {\mathcal O}}$ is well known (refs. below). 
$$
\left\{\begin{array}{l} {t^{*}=\frac{c}{a_{0} } sh\frac{a_{0}t }{c} } \\ {x^{*}=\frac{c²}{a_{0} } \left[ch\left(\frac{a_{0}t }{c} \right)-1\right]} \end{array}\right.  
$$

Now  if we calculate the position of ${\it {\mathcal P}}$ in the  frame of ${\it {\mathcal O}}$ at ${\it {\mathcal O}}'s$ proper time t, we obtain (ref.3): 
$$
x_{\it {\mathcal P}} (t)=\frac{x_{0} +c²/a_{0} }{\cosh (a_{0} t/c)} -\frac{c²}{a_{0} }
$$
 Therefore the velocity of ${\it {\mathcal P}}$ according to ${\it {\mathcal O}}$ is just the derivative (ref.3,4): 
$$
V_{\it {\mathcal P}} (t)=\frac{dx}{dt} =-c\left(\frac{a_{0} x_{0} }{c²} +1\right)\frac{\sinh (a_{0} t/c)}{\cosh ²(a_{0} t/c)}$$
Let's choose a simple example with : 
$c/a_{0} =1year$(corresponding to $a_{0} =9.5m/s²$ ), and  $x_{0} =2light-years$. We obtain: 
$V_{\it {\mathcal P}} (t)=-3c\sinh (t)/\cosh ²(t)$  (1)
Now let's suppose that ${\it {\mathcal O}}$ stops accelerating at some proper time $t_{1} $. From there on, his movement is now inertial. His velocity in K* at $t_{1} $ is given by $v_{{\it {\mathcal O}}} (t_{1} )=ctanh(a_{0} t_{1} /c)$ (refs.).
If we choose for example $t_{1} =(\ln 3)years$ (i.e. $t_{1} \simeq 1.1year$), we obtain: $v_{{\it {\mathcal O}}} (t_{1} )=0.8c$. 
Now, the velocity of ${\it {\mathcal P}}$ relative to ${\it {\mathcal O}}$ at $t_{1} =\ln 3$ , according to equation (1), is $V_{\it {\mathcal P}} (t_{1} )=-1.44c$.
Now comes my question: since after $t_{1} $, the motion of ${\it {\mathcal O}}$ is inertial, the overall situation regarding ${\it {\mathcal O}}$ and ${\it {\mathcal P}}$ is inertial, and therefore, the velocity of ${\it {\mathcal P}}$  relative to ${\it {\mathcal O}}$ is, for $t>t_{1} $, $V_{\it {\mathcal P}} (t)=-v_{{\it {\mathcal O}}} (t)=-v_{{\it {\mathcal O}}} (t_{1} )=-0.8c$ . Therefore, we see that at proper the time $t_{1} $ where ${\it {\mathcal O}}$ stops accelerating, the velocity $V_{\it {\mathcal P}} (t)$ of the inertial object ${\it {\mathcal P}}$ undergoes a discontinuity from (-1.44c) to (-0.8c). 
According to some authors, ``This is totally impossible: the velocity must remain continuous''. 
What should I conclude, and how should I interpret the above results?
References : 


*

*Hobson M.P. \& al. : ``General relativity'', Cambridge UP (2006); p.21

*Rindler W. : ``Relativity'', Oxford UP (2006)

*Adler C.G., Brehme R.W., ``Relativistic solutions for the falling body in a uniform gravitation field'', Am. J. Phys. 59, 209 (1991); III-A

*Gourgoulhon E.,''Special relativity in general frames'', Springer(2010); p.412,413
 A: The coordinate velocity does indeed change discontinuously, but only if the acceleration changes discontinuously i.e. the jerk is infinite. Since for any physical system none of the time derivatives of position can be infinite, in a physical system the coordinate velocity can't change discontinuously. But let's ignore this for now and examine why we get a discontinuous change in coordinate velocity.
Note that I've used the phrase coordinate velocity. In relativity there is an invariant velocity called the four-velocity and this behaves in a generally sane manner for all observers. The coordinate velocity is not an invarient and depends on the choice of coordinate system. What is happening here is not that any invariant physical property is changing discontinuously, but rather that the discontinuous change is in the the coordinate system.
In your example the observers sat on the Earth watching the rocket are using the usual flat space coordinates. However the accelerating observer is using Rindler coordinates. When I say using, I mean that the accelerating observer will observe time dilation and length contraction effects described by the Rindler metric not the Minkowski metric. So the Earth observers and the rocket observer will disagree about velocities. In fact the disagreement eventually becomes extreme because the Rindler coordinates contain a horizon analogous to the event horizon of a black hole. The accelerating observer will observe the Earth initially accelerate away, but as the Earth nears the Rindler horizon the accelerated observer will observe the Earth to slow asymptotically to zero at the horizon. By contrast the Earth observer will see the rocket accelerate asymptotically to $c$.
Your question is asking what happens if the rocket suddenly stops accelerating. The answer is that the observers on the rocket suddenly switch from the Rindler metric to the Minkowski metric, and all coordinate dependant measurements will suddenly switch at the same time. But nothing physical is happening - the changes are just changes in the coordinate system. As I mentioned at the outset, in practice the acceleration can't change discontinuously but would fall smoothly to zero. That means the metric used by the rocket would change smoothly to the Minkowski metric, and all the coordinate dependant measurements would change smoothly as well.
