Suppose the equation of motion of a material point is given by the equation $s(t)=e^t$. Does it follow that $s(t)=v(t)=a(t)=e^t$?

Suppose that the path traveled by a material point changes according to the law $s(t)=e^t$, where $e$ is Euler's number. We know that acceleration is the first derivative of velocity and the second derivative of path, and velocity is the first derivative of path. It is known that the derivative of the exponent is equal to the exponent itself ($f''(x)=f'(x)=f(x)=e^x$). The following equality comes out: $a(t)=v'(t)=s'(t)=(e^t)'=(e^t)'=e^t$. It follows that $a(t)=v(t)=s(t)=e^t$. A purely physical example: at time $t = 1$ s we get that $a(1)=v(1)=s(1) = 1$ $\{ m/s^2; m/s; m \}$.

It turns out that if a material point moves according to the law $s(t)=e^t$, then for any values of $t\to\infty$ the acceleration of the material point equals its same speed and the distance traveled. This is purely mathematical.

Let's apply knowledge of kinematics: $s(t)=0.5at^2$ $(v_0 = 0)$, take advantage of the fact that $s(t)=a(t) \implies s=0.5st^2$, divide both parts of the equation by s and multiply by $2$, we get $2=t^2$, hence we get that $t = \sqrt2 = 1.41$ s. Time cannot be static. So it is purely physically impossible.

The question is, what is it?

Simplified mini example to make it clearer: let's say $s(t)=2^t$, so $v(t)=2^t\ln(2)$, everything is simple - when $t = 3$ s: $s(3)=2^3=8$ m and $v(3)=2^3\ln(2)=5.6$ m/s. Why is it that if $s(t)=2^t$ all is well, but if $s(t)=e^x$ all is impossible?

I do not understand something, or a discrepancy in the mathematical apparatus and physics? I'll ask you to explain in more detail what I'm wrong, most likely I'm wrong.

Of course, I understand that in real life it is difficult to imagine a situation where a material point will move according to the law $s(t)=e^t$, so please just take it as a given.

I apologize for my English, the text was translated into English with the help of a translator.

Your $s=\frac{1}{2} a t^2$ assumes $a$ is constant which it is not here.

The one point that you're actually correct about is that $s(t)=e^t$ is not dimensionally correct. You must actually have $s(t)=s_0 e^{t/t_0}$, where $s_0$ is a constant with length units and $t_0$ is a constant with time units. Then $v$ and $a$ have the right units as you differentiate (since a factor of $t_0^{-1}$ pops out with each derivative).