A particle starts from rest with acceleration $2\mathrm{m}/{\mathrm{s}}^2.$  The acceleration of the particle decreases down to zero uniformly during the time interval of 4 s. The velocity of the particle after 2 s is

Since variation of acceleration is uniform so, $\mathrm{a}=\mathrm{mt}+\mathrm{c}$.

According to the problem, at t = 0, $\mathrm{a}=2\mathrm{m}/{\mathrm{s}}^2.$

$\therefore 2=\mathrm{m}\times 0+\mathrm{c}\Rightarrow \mathrm{c}=2.$

$\mathrm{When} \mathrm{t}=4\mathrm{s},\mathrm{a}=0\therefore 0=\mathrm{m}\times 4+2\Rightarrow \mathrm{m}=-\frac{1}{2}.$

$\therefore \mathrm{a}=-\frac{\mathrm{t}}{2}+2\Rightarrow {\int }_0^{\mathrm{v}}\mathrm{dv}=-\frac{1}{2}{\int }_0^2\mathrm{t}\mathrm{dt}+2{\int }_0^2\mathrm{dt}\Rightarrow \mathrm{v}=3\mathrm{m}/\mathrm{s}.$