Proper Time for Deceleration in Special Relativity As stated here: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
A rocket which accelerates to some point, then decelerates back to 0 over the same distance will measure a proper time for the deceleration as being almost equal to the proper time measured for the positive acceleration portion. It also says that the coordinate time for the deceleration will be far smaller than that of the time during positive acceleration. However, given this equation:
$$\tau = \frac{c}{\alpha}[\operatorname{arsinh}(\frac{\alpha t+w_0}{c}) - \operatorname{arcsinh}   (\frac{w_0}{c})]$$
Where $w_0$ is equal to the initial velocity adjusted for relativity (the velocity at which it begins to decelerate) and $t$ is the coordinate time for the deceleration, the equation returns a proper time for the deceleration that is significantly smaller than that of the positive acceleration due to the significantly smaller coordinate time. Specifically in reference to when he says: 

If you want to arrive at your destination and stop then you will have to turn your rocket around half way and decelerate at 1g. In that case it will take nearly twice as long in terms of proper time T for the longer journeys; the Earth time t will be only a little longer, since in both cases the rocket is spending most of its time at a speed near that of light. 

Is Mr. Koks wrong or am I missing something conceptually? It seems mathematically impossible to have the proper time for the deceleration to be almost as long as the proper time for the positive acceleration due to the significantly smaller coordinate time.
 A: 
Is Mr. Koks wrong or am I missing something conceptually?

You're simply misunderstanding him.
Here is the full context of article in question:

These equations are valid in any consistent system of units such as
  seconds for time, metres for distance, metres per second for speeds
  and metres per second squared for accelerations.  In these units c = 3
  × 108 m/s (approx).  To do some example calculations it is easier to
  use units of years for time and light years for distance.  Then c = 1
  lyr/yr and g = 1.03 lyr/yr2.  Here are some typical answers for a =
  1g.
T          t         d          v                γ
1 year    1.19 yrs  0.56 lyrs  0.77c                 1.58  
2         3.75      2.90       0.97                  3.99
5        83.7      82.7        0.99993              86.2
8     1,840     1,839          0.9999998         1,895    12   113,243   113,242          0.99999999996   116,641

So in theory you can travel across the galaxy in just 12 years of your
  own time.  If you want to arrive at your destination and stop then you
  will have to turn your rocket around half way and decelerate at 1g. 
  In that case it will take nearly twice as long in terms of proper time
  T for the longer journeys; the Earth time t will be only a little
  longer, since in both cases the rocket is spending most of its time at
  a speed near that of light.  (We can still use the above equations to
  work this out, since although the acceleration is now negative, we can
  "run the film backwards" to reason that they still must apply.)

Mr. Koks is saying that the total proper time, for a journey that constantly accelerates half the distance and then constantly decelerates (at the same rate) the remaining half, will be nearly twice as long as a journey that accelerates at a constant rate the entire distance.
Once the rocket reaches ultra-relativistic speed, the proper time nearly stops.  So, it is reasonable that the proper is nearly twice as long for a journey that decelerates to sub-relativistic speed and then to a stop at its destination instead of speeding by it at near light speed.
Also, it is reasonable that the total coordinate time does not change that much between the two scenarios for the reason give by the author. Do you see why?
A: The Koks article covers a lot of different stuff, not just the kinematics that you're asking about. For a clearer description of the kinematics, try this WP article.
The world-line of a rocket with constant acceleration is a hyperbola in the rest frame.
The rocket trip with turn-around would look like an S-shaped curve, formed by gluing together two pieces of such a hyperbola. The shape is perfectly symmetric, so the coordinate time is exactly equal for the two legs of the trip, and likewise for the proper time. This is all for the case where you accelerate with constant proper acceleration $\alpha$ from a velocity of $0$ to some maximum speed $v$, then decelerate with $-\alpha$ back down to $0$ again. Is that what you had in mind, or something more complicated?

Where $w_0$ is equal to the initial velocity adjusted for relativity and $t$ is the coordinate time for the deceleration[...]

I'm not sure what you mean by "adjusted for relativity." And isn't the initial velocity $0$? Does the notation $w_0$ occur somewhere in the Koks article? I don't see it.
A: Alfred beat me to it while I was preparing my answer, but I'll post it anyway :-)
Mr. Koks is comparing two situations: case 1), where the rocket has a constant acceleration during the entire trip, and case 2), where the rocket accelerates half the trip, and then decelerates again during the second half. Let's compare the total proper time and coordinate time in these two cases.
First of all, the velocity of a rocket with constant positive proper acceleration $g$ and initial velocity $v_0=0$ is
$$
v(t) = \frac{gt}{\sqrt{1 + g^2t^2/c^2}}.
$$
Integration gives us the distance $x$ travelled by the rocket 
$$
x = \frac{c^2}{g}\left(\sqrt{1 + g^2t^2/c^2}-1\right).
$$ 
Also, let's call $t_\max$ the coordinate time at which the velocity of the rocket is at its maximum, and call $D$ the total distance travelled by the rocket.
Case 1)
When there is no deceleration, the rocket has its maximum velocity at the destination $x=D$, so we get from the equation above
$$
t_\max = \sqrt{\frac{D^2}{c^2} + \frac{2D}{g}}.
$$
The total elapsed proper time $\tau_1$ then equals
$$
\tau_1 = \frac{c}{g}\text{arcsinh}\left(\frac{g\,t_\max}{c}\right)
= \frac{c}{g}\ln\left(\frac{g\,t_\max}{c}+\sqrt{1 + \frac{g^2\,t^2_\max}{c^2}}\right),
$$
and the total elapsed coordinate time $t_1$ is of course
$$
t_1 = t_\max = \sqrt{\frac{D^2}{c^2} + \frac{2D}{g}}.
$$
Now, suppose that $D$ is very large. In that case,
$$
\begin{align}
t_1 &= t_\max \approx D/c,\\
\tau_1 &\approx \frac{c}{g}\ln\left(\frac{2g\,t_\max}{c}\right)
\approx \frac{c}{g}\ln\left(\frac{2Dg}{c^2}\right) = \frac{c}{g}\ln\left(\frac{Dg}{c^2}\right) + \frac{c}{g}\ln 2 \approx\frac{c}{g}\ln\left(\frac{Dg}{c^2}\right).
\end{align}
$$
Case 2)
If the rocket accelerates during the first half and decelerates during the second half, then it will have its maximum velocity at the midpoint $x=D/2$, so that
$$
t'_\max = \sqrt{\frac{D^2}{4c^2} + \frac{D}{g}}.
$$
Now, the situation during the second half is completely symmetrical to the first half of the trip: it will take the rocket just as long to decelerate as it took to accelerate, so that the total elapsed coordinate time $t_2$ is
$$
t_2 = 2t'_\max = \sqrt{\frac{D^2}{c^2} + \frac{4D}{g}},
$$
and the total elapsed proper time $\tau_2$ is
$$
\tau_2 = \frac{2c}{g}\text{arcsinh}\left(\frac{g\,t'_\max}{c}\right).
$$
For large $D$, we get
\begin{align}
t_2 & \approx D/c,\\
\tau_2 &\approx\frac{2c}{g}\ln\left(\frac{Dg}{c^2}\right).
\end{align}
So we see that $t_2$ is of the same order as $t_1$, while $\tau_2$ is about twice as large as $\tau_1$.
