\textbf{Exercise 4 — Candidates who have not followed the specialization course}

We are interested in the fall of a water droplet that detaches from a cloud without initial velocity. A very simplified model makes it possible to establish that the instantaneous vertical velocity, expressed in $\mathrm{m.s^{-1}}$, of the droplet's fall as a function of the fall duration $t$ is given by the function $v$ defined as follows:

For every non-negative real number $t$, $v(t) = 9.81\dfrac{m}{k}\left(1 - \mathrm{e}^{-\frac{k}{m}t}\right)$; the constant $m$ is the mass of the droplet in milligrams and the constant $k$ is a strictly positive coefficient related to air friction.

We recall that instantaneous velocity is the derivative of position.\\
Parts $A$ and $B$ are independent.

\textbf{Part A - General case}
\begin{enumerate}
  \item Determine the variations of the velocity of the water droplet.
  \item Does the droplet slow down during its fall?
  \item Show that $\lim_{t \rightarrow +\infty} v(t) = 9.81\dfrac{m}{k}$. This limit is called the terminal velocity of the droplet.
  \item A scientist claims that after a fall duration equal to $\dfrac{5m}{k}$, the velocity of the droplet exceeds $99\%$ of its terminal velocity. Is this claim correct?
\end{enumerate}

\textbf{Part B}

In this part, we take $m = 6$ and $k = 3.9$.\\
At a given instant, the instantaneous velocity of this droplet is $15\mathrm{~m.s^{-1}}$.
\begin{enumerate}
  \item How long ago did the droplet detach from its cloud? Round the answer to the nearest tenth of a second.
  \item Deduce the average velocity of this droplet between the moment it detached from the cloud and the instant when its velocity was measured. Round the answer to the nearest tenth of $\mathrm{m.s^{-1}}$.
\end{enumerate}