Of course, $e$ is ubiquitous in kinematics. For example, consider a repulsive force proportional to $x$, $$F = kx.$$ Then the acceleration is $$a = \frac{k}{m} x = \omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}.$$ This differential equation has solutions of the form $e^{\omega t}$ and $e^{- \omega t}$. In particular, suppose that $x(0) = 1$ and $v(0) = \omega$. In that case the solution is exactly $x(t) = e^{\omega t}$, so $$x(1) = e.$$ In general, for any linear force law, the solutions will be exponentials or complex exponentials, so it's honestly hard to avoid using $e$.

Of course, $e$ is ubiquitous in kinematics. For example, consider a repulsive force proportional to $x$, $$F = kx.$$ Then the acceleration is $$a = \frac{k}{m} x = \omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}.$$ This differential equation has solutions of the form $e^{\omega t}$ and $e^{- \omega t}$. In particular, suppose that $x(0) = 1$ and $v(0) = \omega$. In that case the solution is exactly $$x(t) = e^{\omega t}.$$ In general, for any linear force law, the solutions will be exponentials or complex exponentials, so it's honestly hard to avoid using $e$.