(d) $\frac{{dv}}{{dt}} = 6 – 3v \Rightarrow \frac{{dv}}{{6 – 3v}} = dt$

Integrating both sides, $\int {\frac{{dv}}{{6 – 3v}}} = \int {dt} $

$⇒ \frac{{{{\log }_e}(6 – 3v)}}{{ – 3}} = t + {K_1}$

$⇒ {\log _e}(6 – 3v) = – 3t + {K_2}$…(i)

At $t = 0,\;v = 0$

$\therefore {\log _e}6 = {K_2}$

Substituting the value of ${K_2}$ in equation (i)

${\log _e}(6 – 3v) = – 3t + {\log _e}6$

$⇒ {\log _e}\left( {\frac{{6 – 3v}}{6}} \right) = – 3\,t$ $⇒$ ${e^{ – 3t}} = \frac{{6 – 3v}}{6}$

$⇒ 6 – 3v = 6{e^{ – 3\,t}}$ $⇒$ $3v = 6(1 – {e^{ – 3\,t}})$

$⇒ v = 2(1 – {e^{ – 3\,t}})$

$\therefore {v_{{\rm{terminal}}}} = 2\;m/s$  (When $t = \infty $).

Acceleration $a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ {2\left( {1 – {e^{ – 3\;t}}} \right)} \right] = 6{e^{ – 3\,t}}$

Initial acceleration =$6\;m/{s^2}$.