Question

a= 3-6v find the velocity of particle at t= 2 second and particle start from rest?​

Answer 1

Answer:

The original question is a bit confusing as it implies that displacement and distance is the same thing, which it is not.

I have set up the necessary integration for each different case hereunder.

Explanation:

Total distance (scalar quantity representing actual path length) is given by the sum of the partial integrals 

x=∫1−3(0−(−t2+3t−2)dt+∫21(−t2+3t−2)dt+∫62(t2−3t+2)dt

Total displacement (vector quantity representing straight line drawn from start to end of motion) is given in magnitude by the following integral

∣∣→x∣∣=−∫1−3(t2−3t+2)dt+∫21(−t2+3t−2)dt−∫62(t2−3t+2)dt

The graph of the velocity function with time makes it clear why these integrals need to be set up for the vector rules to be obeyed and the definitions to be satisfied. 

graph{-x^2+3x-2 [-34.76, 38.3, -21.53, 14.98]}