The retardation fo a moving particle if the relation between time and position is ` t= A x^3 + Bx^2 ` where ` A` and ` B` are appropriate constants will be .

Explanation

Give, ` t= Ax^3 = Bx^2` ltbgt :. ` (dt)/(dx) =3 Ax^2 = 2 2 Bx`
or ` v= (dx)/(dt) = ( 3 Ax^2 + 2 Bx)^(-10`
Now , (dv) /dx) = (-1) (3 Ax^2 + 2 Bx)^(-1)`
Acceleration , ` a = 9dv)/(dt0 = (dv)/(dx) xx (dx)/(dt)`
` = (-(6 Ax^2 + 2 B))/((3 A x^2 + 2 Bx)^20 xx 1/( (3 Ax^2 + 2 Bx0`
`= (-(6 Ax+ 2 B))/( (2 Ax^2 = 2 Bx)^3)`
Retardation ` =- a = ( 6 Ax+ 2 B)/((3 Ax^2 + 2 Bx)^3)`.