Hello there!

Greetings!


It is given that the displacement is :-

x^2 = at^2 + b.

On differentiating it with respect to time we get :-

2xdx = 2atdt + 0

On 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)

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

dx/dt = at/(at^2 + b) ..............( 2)

Now, once 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 eqns 3& 4, acceleration is proportional to 1/x^3

Thankyou