For a moving paritcle, the relation between time and position is given by t = Ax ^(2) + Bx. Where A and B are contants. Find the acceleration of the particle as a function of velocity.

For a moving paritcle, the relation between time and position is given

1.	For a moving paritcle, the relation between time and position is given by t = Ax ^(2) + Bx. Where A and B are contants. Find the acceleration of the particle as a function of velocity.
Answer» SOLUTION :Here, `t = Ax ^(2) + Bx` is given `therefore (dt)/(DX) = 2 A x +B` `therefore v = (dx)/(dt) = (2Ax +B) ^(-1) ""…(1)` Now, acceleration, `a = (dv)/(dt)= (dv)/(dx) xx (dx)/(dt) (because "Multiply by" (dx)/(dx))` `= [(d)/(dx) (2 A x + B) ^(-1) ](v) ` (From eq. (1)) `= (-1 ) (2A) (2 Ax + B) ^(-2) (v)` But if ` (2A x + B) ^(-1)=v,` then `(2Ax +B) ^(-2) = v ^(2) ` `therefore a =- 1A v ^(3)`