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.

Explanation

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)`