A body has an initial velocity of 3 m/s and has an acceleration of 1 m/sec2 normal to the direction of the initial velocity. Then, its velocity 4 seconds after the start, is:

ux=3m/s,ax=0uy=0, ay=1m/sec2 and t=4secIf vx and vy be the velocities after 4 sec respectively, thenvx=ux+axt=3ms−1vy=uy+ayt=0+1×4=4ms−1v=vx2+vy2=32+42=9+16=5m/sAngle made by the resultant velocity w.r.t. direction of initial velocity, i. e.,x-axis, isβ=tan−1⁡vyvx=tan−1⁡43