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 60o 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 getv1=ucos⁡60∘+at=2×12+2×2=5m/sVelocity perpendicular to direction of acceleration remains same, which is usin⁡60∘=3m/sSo net velocity at the end of 2 s:v=v12+(3)2=52+3=28=27m/s