the displacement of particle is given byX= a_(0) + (a_(1)t)/(2)-(a_(2)t^(2))/(3). What is its acceleration?

the displacement of particle is given byX= a_(0) + (a_(1)t)/(2)-(a_(2)

1.	the displacement of particle is given byX= a_(0) + (a_(1)t)/(2)-(a_(2)t^(2))/(3). What is its acceleration?
Answer» `(2a_(2))/(3)` `-(2a_(2))/(3)` `a_(2)` zero Solution :Displacement `X = (a_(0)+a_(1)t)/(2) - (a_(2)t^(2))/(3)` VELOCITY `V = (dx)/(DT)` `= (d)/(dt) (a_(0) +(a_(1)t)/(2)-(a_(2)t^(2))/(3))` `v = (a_(1))/(2)- (2a_(2)t)/(3)` Acceleration `a = (dv)/(dt) = (d)/(dt) ((a_(1))/(2) - (2a_(2)t)/(3))` `= (-2a_(2))/(3)`