The motion of a body is given by the equation $\frac{dv\left(t\right)}{dt}=8-4v\left(t\right)$, where $v\left(t\right)$ is speed in $m/s$ and t in sec. If the body was at rest at $t=0$ , then

$\frac{dv}{dt}=8-4v\Rightarrow \frac{dv}{8-4v}=dt$

$\int \frac{dv}{8-4v}=\int dt$$\Rightarrow \frac{{\log }_e\left(8-4v\right)}{-4}=t+k_1$

$\Rightarrow {\log }_e\left(8-4v\right)=-4t+k_2$

at   $t=0,v=0,$ ${\log }_e^8=k_2$

${\log }_e^{\left(8-4v\right)}=-4t+{\log }_e^8$

$\Rightarrow \log \frac{\left(8-4v\right)}{8}=-4t\Rightarrow e^{-4t}=\frac{8-4v}{8}$

$-8e^{-4t}+8=4v\Rightarrow 4v=8\left(1-e^{-4t}\right)$

$v=2\left(1-e^{-4t}\right)$ option (B) correct

$v_{ter\min al}=2m/s$ (when $t=\alpha$) option (A) correct

Acceleration   $a=\frac{dv}{dt}=\frac{d}{dt}\left(2\left(1-e^{-4t}\right)\right)=8e^{-4t}$

Initial acceleration $=8m/s^2$ option (D) correct

When acceleration is half the initial value i.e. $a=8e^{-4t}$

$4=8e^{-4t}\Rightarrow e^{-4t}=\frac{1}{2}\Rightarrow v=2\left(1-\frac{1}{2}\right)=1m/s$