The motion of a body is given by the equation $\frac{dv}{dt}=6-3v$, where $v$ is speed in $m/s$ and $t$ is time in seconds. If the body is at rest at $t=0$, then

The correct options are
A the terminal speed is $2m/s$.
B the speed varies with the time as $v=2(1-e^{-3t})m/s$.
D the magnitude of the initial acceleration is $6m/s^2$.
Given, $\frac{dv}{dt}=6-3v$
Re- writing it as, $\frac{dv}{2-v}=3dt$.
Given $v=0$ at $t=0$. Integrating both sides
${\int }_0^v\frac{dv}{2-v}=3{\int }_0^{t}dt$,
to get
$v=2(1-e^{-3t}).....\left(1\right)$
Thus, terminal velocity of the body is
$v_t=\underset{t\to \infty }{lim}2(1-e^{-3t})=2m/s$
The acceleration of the body is
$a=\frac{dv}{dt}=\frac{d}{dt}\left(2\right(1-e^{-3t})=6e^{-3t}$
This gives initial acceleration as $6m/s^2$.
Graphs plotted of $a-t,v-t$ and $s-t$ are shown as below

When acceleration is half the initial value i.e $a=3m/s^2$ , then
$3=6e^{-3t}$
$\Rightarrow e^{-3t}=\frac{1}{2}$
Put this in equation $\left(1\right)$ to get velocity at this instant of time
$\Rightarrow v=1m/s^2$
Hence, option (c) is incorrect.