Mass (m) = 3 kg, force (F) = $6{t^2}\widehat i + 4t\widehat j$

$\therefore$ acceleration
$a = F/m = {{6{t^2}\widehat i + 4t\widehat j} \over 3} = 2{t^2}\widehat i + {4 \over 3}t\widehat j$

Now, $a = {{dv} \over {dt}} = 2{t^2}\widehat i + {4 \over 3}t\widehat j$

$dv = \left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)dt$

$\therefore v = \int\limits_0^3 {\left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)} dt$

= $\left. {{2 \over 3}{t^3}\widehat i + {4 \over 6}{t^2}\widehat j} \right|_0^3 = 18\widehat i + 6\widehat j$