Equation of motion of a body is $\frac{dv}{dt}=-4v+8$, where $v$ is the velocity in ${\mathrm{ms}}^{-1}$ and $t$ is the time in second. Initial velocity of the particle was zero. Then,

$\frac{dv}{dt}=a=-4v+8$
$\therefore \frac{da}{dt}=-4\frac{dv}{dt}=-4(-4v+8)$
$=16 v-32$
$\therefore {\left(\frac{da}{dt}\right)}_t=16v_i-32$
$=\left(16\right)\left(0\right)-32$
$=-32 \mathrm{m}/{\mathrm{s}}^2$
Further, ${\int }_0^y\frac{dv}{8-4v}={\int }_0^{t}dt$
Solving this equation we get,

$v=2(1-e^{-4t})$
Hence, $v-t$ graph is exponentially increasing graph, terminating at $2 \mathrm{m}/\mathrm{s}$.