d
(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 =\(6\;m/{s^2}\).