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,
$$y = k - (k - v) \cdot e^{-rx}$$