Given the displacement as :-

x^2 = at^2 + b.

Differentiating it with respect to time we get :-

2xdx = 2atdt + 0

Rearranging the terms we have :-

dx/dt = at/x  (dx/dt is velocity)  .... 1

Now x from the given equation can be written as x = √(at^2 + b)

Substituting value of x from above to equation number 1 we get :-

dx/dt = at/(√at^2 + b) ...... 2

Now again differentiating equation 2 with respect to time (derivative of velocity is acceleration) we get :-

dv/dt = A( acceleration )

{ (√at^2 +b)*a - a^2*t^2/(√at^2 +b) }/ (at^2 + b)

Therefore , A = ab/[(at^2 + b)^3/2] .... 3

Now since x = √(at^2 + b) ....4

From equation 3 and 4 Acceleration is proportional to 1/(x^3).

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