The relativistic generalization of "uniform acceleration" is a timelike curve of constant curvature in Minkowski spacetime.

One should use really use rapidities [the analogue of the angle].

$v=u+at$ is generalized to $\theta=\theta_0+\frac{\rho}{c} t$ (where $\rho$ is the proper-acceleration), which can be expressed in terms of velocity by using $v=c\tanh\theta$. So, we have $$v=c\tanh\theta=c\tanh(\theta_0+\frac{\rho}{c} t)=c\frac{(v_0/c)+\tanh(\frac{\rho}{c} t)}{1+(v_0/c)\tanh(\frac{\rho}{c} t)}.$$

One would have to integrate that out to get the position function. From my notes [and my rushed attempt to restore the c's], I get $$y-y_0=\frac{c^2}{\rho}\left( \sqrt{1+\left(\frac{\rho}{c} t+\gamma \frac{v_0}{c}\right)^2}-\gamma \right),$$ where $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$.

The relativistic generalization of "uniform acceleration" is a timelike curve of constant curvature in Minkowski spacetime.

One should use really use rapidities [the analogue of the angle].

$v=u+at$ is generalized to $\theta=\theta_0+\frac{\rho}{c} t$ (where $\rho$ is the constant proper-acceleration), which can be expressed in terms of velocity by using $v=c\tanh\theta$. So, we have $$v=c\tanh\theta=c\tanh(\theta_0+\frac{\rho}{c} t)=c\frac{(v_0/c)+\tanh(\frac{\rho}{c} t)}{1+(v_0/c)\tanh(\frac{\rho}{c} t)}.$$

One would have to integrate that out to get the position function. From my notes [and my rushed attempt to restore the c's], I get $$y-y_0=\frac{c^2}{\rho}\left( \sqrt{1+\left(\frac{\rho}{c} t+\gamma_0 \frac{v_0}{c}\right)^2}-\gamma_0 \right),$$ where $\gamma_0=\frac{1}{\sqrt{1-(v_0/c)^2}}=\cosh\theta_0$.