The motion of a body is given by the equation  $\frac{dv\left(t\right)}{dt}=6.0-3v\left(t\right)$. where  $v\left(t\right)$ is speed in m/s and  t in sec . If body was at rest at t=0 

 $\frac{dv}{dt}=6-3v\Rightarrow \frac{dv}{6-3v}=dt$
Integrating both sides,  $\int \frac{dv}{6-3v}=\int dt$
 $$\begin{gathered} \Rightarrow \frac{{\log }_e(6-3v)}{-3}=t+K_1 \\ \Rightarrow {\log }_e(6-3v)=-3t+K_2-------\left(i\right) \end{gathered}$$
At $t=0,v=0\therefore {\log }_{e}6=K_2$
Substituting  the value of  $K_2$ in equation (i)
${\log }_e(6-3v)=-3t+{\log }_{e}6$ 
 $$\begin{gathered} {\log }_e\left(\frac{6-3v}{6}\right)=-3t\Rightarrow e^{-3t}=\frac{6-3v}{6} \\ \Rightarrow 6-3v=6e^{-3t}\Rightarrow 3v=6(1-e^{-3t}) \\ \Rightarrow v=2(1-e^{-3t}) \end{gathered}$$

$v\text{ terminal =}2\text{ m/s (when t=}\infty \text{)}$
 Acceleration  $a=\frac{dv}{dt}=\frac{d}{dt}\left[2(1-e^{-3t})\right]=6e^{-3t}$
Initial acceleration = $6{\text{ m/s}}^2$.