$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 $