The equation $F = m v$ is not time-symetrical, i.e. reversing the sign of $t$ gives a different equation because of the appearence of velocity. The equation $F = m a$ is second order in $t$, so it doesn't change under time inversion $t \Rightarrow -\, t$, but only if the left member ($F$) doesn't include terms with dependance on velocity (like friction). If there's a friction, then $F$ contains a non-symetrical term under time-inversion. Friction implies a loss of information.
If you have a strict exponential damping: $x(t) = x_0 \, e^{- \lambda t}$, then measuring position and velocity at time $t_1$ gives this system of equations:
\begin{align}
x_1 &= x_0 \, e^{- \lambda t_1}, \tag{1} \\[12pt]
v_1 &= -\, \lambda \, x_1. \tag{2}
\end{align}
You then knows $x_1$ and $v_1$ (so you know $\lambda$ from (2)), but you don't know time $t_1$ and want to retrodict the initial position $x_0$ (at time $t_0 = 0$). Equation (1) gives you only one equation for two unknows. You need more information to retrodict the past. Measuring position and velocity again at time $t_2$ gives you two new equations:
\begin{align}
x_2 &= x_0 \, e^{- \lambda t_2}, \tag{3} \\[12pt]
v_2 &= -\, \lambda \, x_2. \tag{4}
\end{align}
Equation (4) is useless. Combining (1) and (3) gives this, where $\Delta t = t_2 - t_1$ is known:
\begin{equation}\tag{5}
x_2 = x_0 \, e^{- \lambda (t_1 \,+\, \Delta t)} = x_1 \, e^{- \lambda \, \Delta t},
\end{equation}
so there's nothing new here, and you can't find $x_0$!
In other words: The pure exponential function has no memory! This fact is very important for the radioactive decay and for statistical theory.
