A body of mass 1 kg moving with an initial speed of $v_0=9\mathrm{m}/\mathrm{s}$ starts experiencing a variable resistive force directed opposite its velocity. Let v denote the speed of the body at time t and F be the magnitude of the resistive force. If F varies as $F=2\sqrt{v}$ then variation of F with time is  given by

The 1-kg body experiences an acceleration  $a=\frac{-F}{1\mathrm{kg}}=-2\sqrt{v}$ due to the resistive force. Writing  $v\frac{dv}{dx}$ in place of c, we have

$\begin{array}{l}v\frac{dv}{dx}=-2\sqrt{v}\\ \Rightarrow \int \sqrt{v}dv=-2\int dx\\ \Rightarrow v^{3/2}=-3x+{\text{ const }}^{27}[v=9\mathrm{m}/\mathrm{s}\text{ at }x=0]\end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v=(3x-27{)}^{2/3}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\int \frac{dx}{(3x-27{)}^{2/3}}=\int dt\end{array}$

$\begin{array}{l}\Rightarrow (3x-27{)}^{1/3}=t+{\mathrm{const}}^{-3}[x=0\text{ at }t=0]\\ \Rightarrow x=9+\frac{1}{3}(t-3{)}^3\end{array}$

Now differentiating twice with respect to time, we have

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a=2(t-3)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}F=-\left(1\mathrm{kg}\right)a=2(3-t)\end{array}$