How can acceleration in special relativity be uniform? Acceleration is defied as the rate of change in velocity, implying that $v(t) = at + v_0$. Say that an object is accelerating at $5 m/s^2$ with respect to an inertial frame in special relativity. Using the definition of acceleration alone, we now have $v(t) = 5t + v_0$. This means that given enough time, $v(t) > c$ which is supposed to be impossible in relativity. The solution is to say that uniform acceleration in special relativity is hyperbolic, so while it may approach c it never actually gets there. The problem with this is how can that be said to be uniform acceleration? If something is accelerating at $5m/s^2$ in order for it to stay less than $c$ it has to eventually decelerate even more and more to asymptotically approach $c$.
How is hyperbolic acceleration uniform?
 A: I’ll just formalize my previous comments. Let me restrict to 2D flat spacetime with a certain inertial frame $t,x$ ($c=1$ and the metric signature is $(+,—)$ like in particle physics). Then hyperbolic motion of proper acceleration $a$ can be parametrized by its proper time $\tau$ given (up to a space-time translation):
$$
t = \frac{\sinh(a\tau)}{a} \\
x = \frac{\cosh(a\tau)}{a} \\
$$
As you pointed out, the acceleration viewed from the original frame is not uniform:
$$
x = \frac{\sqrt{1+(at)^2}}{a} \\
\frac{dx}{dt} = \frac{at}{\sqrt{1+(at)^2}}\\
\frac{d^2x}{dt^2} = -a\frac{1}{\sqrt{1+(at)^2}^3}
$$
the velocity reaches $1$ asymptotically, so you have a non uniform deceleration reaching $0$ asymptotically.
Note however that acceleration $d^2x/dt^2$ in the frame coincides exactly with the proper acceleration $a$ at $t=0$ ie when the frame coincides with the rest frame of the particle.
This is true in general. At any event of the world-line, I can choose an inertial frame which coincides with the rest frame of the particle at that event. The acceleration measured in these identical frames at this specific event will coincide with proper acceleration. This is an equivalent definition of proper acceleration. And it is this proper acceleration that is constant in hyperbolic motion.
Geometrically, this instantaneous rest frame of the particle is the Minkowski analogue of the Frenet basis in the Euclicdean plane. This is why proper acceleration is the analogue of curvature and given generally by:
$$
a=\frac{d^2x}{d\tau^2}\frac{dt}{d\tau}-\frac{d^2t}{d\tau^2}\frac{dx}{d\tau}
$$
which you can check explicitly for hyperbolic motion.
Hope this helps.
