A particle moves on a circular path of radius x of centre O(0,0) with constant speed v=20m/s from a point A(0,-x) to a point B such that OB makes an angle of 30° with x-axis. Calculate the change in velocity from A to B( Please illustrate it using Parallelogram law of vector addition)

The question has not specified whether the motion is clockwise or anticlockwise. Well, for clockwise the answer would be as follows. Intially, when particle is at (0,-x) the velocity vector is perpendicular to x axis and along the positive y-axis. So v1=0 i+v j When the radius vector makes 30 degree angle with x axis, the velocity vector makes 150 degrees angle with the initial vector. SInce vectors are free you can write the final vector as: v2=v(cos(-60)i+sin(-60)j) (Since angle made is -60 with the x axis. So v2=0.5v i - 0.8666v j So delta v=v2-v1 =0.5v i -1.8666v j If u use parallelogram law, then delta v =v2+(-v1) -v1 makes an angle 30 degrees with v2 so delta v =v2v²+2v²cos(30) =v(v(2+v3))