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),

$$\frac{dy}{k-y}=rdx$$

Integrating both sides,

$$\begin{gathered} \int \frac{1}{k-y}dy=\int rdx \\ -\ln (k-y)=rx+C \\ \ln (k-y)=C-rx \\ k-y=e^{C-rx} \\ y=k-e^{\ln c-rx} \\ y=k-c{e}^{-rx} \end{gathered}$$

At x = 0 (if v is the velocity the player initally had),

$$\begin{gathered} k-c=v \\ c=k-v \end{gathered}$$

This assumption is based on the inequality

$$0\le v\le k$$

Substituting this in our equation,

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