If *r* is the rate of growth of speed, *k* is the top speed and *y* is the current speed, $$\frac{dy}{dx} = r (k - y)$$ Multipying both sides by *dx / (k - y)*, $$\frac{dy}{k - y}=r \, dx$$ Integrating both sides, $$\int\frac{1}{k - y} \, dy = \int r \, dx \\ -\ln(k - y) = rx + C \\ \ln(k - y) = C - rx \\ k - y = e^{C - rx} \\ y = k - e^{\ln{c} - rx} \\ y = k - c e^{-rx}$$ At *x = 0* (if *v* is the velocity the player initally had), $$k - c = v \\ c = k - v$$ This assumption is based on the inequality $$0 \le v \le k$$ Substituting this in our equation, $$y = k - (k - v) \cdot e^{-rx}$$