$a=4i$ $\dfrac{dv}{dt}=4i$ $dv=\left(4i\right)dt$ $\int_{v_0}^vdv=\left(4i\right)\int_0^tdt$ $v-v_0=\left(4i\right)t$ $\therefore{}\ {v=v}_0+\left(4t\right)i$ $=i+1.732j+4t.i$ $\therefore{}\ \bar{v}=\left(1+4t\right)i+1.732j$ $at\ t=1\ s;\ \ \bar{v_1}=5i+1.732j$ $\therefore{}\ v_1=\sqrt{5^2+{\left(1.732\right)}^2}=5.29\dfrac{m}{s}$ ${\theta} = {tan}^{-1}{\dfrac{1.732}{5}}=19.11$