kinematics - Can the initial time be non-zero?

I am just slightly confused about how initial time is chosen in mechanics. I keep thinking I have understood it but some doubts come up later.

Basically when I derived the constant acceleration equations in a general way I ended up with $v(t)=v(0)+at$ and $x(t)=x(0)+v(0)t+\frac{1}{2} at^2$. However, I didn't set any initial conditions at this point, all I did was derive them using integration then set $t=0$ to get the constants of integration.

Now I understand that we can use a value other than $t=0$ to find the constants, however $t=0$ is convenient.

What I don't understand, is whether by setting $t=0$ to find the constants, I have now somehow implicitly chosen $t=0$ as the initial time? Because if $t=0$ was not the initial time, then it should be impossible to have $t=0$ (am I wrong in thinking this?)

In short, my question is: did I implicitly choose an initial time of $t=0$? If not, then regardless of how ugly the end result becomes, am I free to choose a nonzero initial time so that my first equation becomes $\Delta v = v(0)+at-v(t_0)$ (where I simply subtracted $v(t_0)$ from both sides, where $t_0\ne 0$, as I just said) and the second one becomes $\Delta x = x(0)+v(0)t+\frac{1}{2} at^2-x(t_0)$. I am not concerned with how ugly these are, I am simply wondering if they are correct in principle?

Also, if I wanted to make them not-ugly, I know how to do that: I would choose the $t$ value when finding the constants of integration as $t=t_0$ rather than $t=0$. In other words, my initial time is $t=t_0\ne 0$, hence $v(t)=\int a dt=at+c$ and instead of setting $t=0$ here, I instead let $t=t_0$ to get $v(t_0)=at_0+c$ or $c=v(t_0)-at_0$, hence $v(t)=v(t_0)+a(t-t_0)$. Then I would integrate again to get $x(t)=v(t_0)t+\frac{1}{2} a(t-t_0)^2+c$ and setting $t=t_0$ again gives $c=x(t_0)-v(t_0) t_0$ so that $x(t)-x(t_0)=v(t_0) (t-t_0)+\frac{1}{2} a(t-t_0)^2$.