Velocity of a particle at time t = 0 is 2 m/s. A constant acceleration of 2 m/s² act on the particle for 2 s at an angle 60^o with the initial velocity. The magnitude of velocity of particle at the end of 2 s will be 

Let us make the components of initial velocity in the direction and perpendicular to the direction of acceleration as shown in the figure.
In the direction of acceleration:

Using $v=u+at$, we get

$\begin{array}{l}{v}_1=u\cos {60}^{\circ }+at\\ =2\times \frac{1}{2}+2\times 2=5\mathrm{m}/\mathrm{s}\end{array}$

Velocity perpendicular to direction of acceleration remains same, which is 

$u\sin {60}^{\circ }=\sqrt{3}\mathrm{m}/\mathrm{s}$

So net velocity at the end of 2 s:

$v=\sqrt{v_1^2+(\sqrt{3}{)}^2}=\sqrt{5^2+3}=\sqrt{28}=2\sqrt{7}\mathrm{m}/\mathrm{s}$