The distance x covered by a particle in one-dimensional motion varies with time t as ${x^2} = a{t^2} + 2bt + c$. The acceleration of the particle varies as:
$
  (a){\text{ }}{{\text{x}}^{ - 3}} \\
  (b){\text{ }}{{\text{x}}^{\dfrac{3}{2}}} \\
  (c){\text{ }}{{\text{x}}^2} \\
  (d){\text{ }}{{\text{x}}^{\dfrac{{ - 2}}{3}}} \\
 $

Hint: In this question use the concept that the velocity is considered as the rate change of position with respect to time that is $v = \dfrac{{dx}}{{dt}}$ and the acceleration is defined as the rate change of velocity with respect to time that is $a = \dfrac{{dv}}{{dt}}$. This will help approaching the problem.

Complete step-by-step answer:

As we know that the velocity of any particle is rate of change of distance covered w.r.t time and acceleration is rate of change of velocity w.r.t. time.
Now it is given that the distance x is covered by a particle in 1-D motion w.r.t. time (t) is given as

${x^2} = a{t^2} + 2bt + c$

Now take square root on both sides we have,

$ \Rightarrow x = \sqrt {a{t^2} + 2bt + c} $..................... (1)
So the velocity (V) of the particle is
$ \Rightarrow v = \dfrac{d}{{dt}}x$ m/s.................... (2)
$ \Rightarrow v = \dfrac{d}{{dt}}\sqrt {a{t^2} + 2bt + c} $
Now differentiate it according to property $\dfrac{d}{{dt}}\sqrt {{t^2}} = \dfrac{1}{{2\sqrt {{t^2}} }}\dfrac{d}{{dt}}{t^2},\dfrac{d}{{dt}}{t^n} = n{t^{n - 1}}$ so use this property in above equation we have,

$ \Rightarrow v = \dfrac{d}{{dt}}\sqrt {a{t^2} + 2bt + c} = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }}\dfrac{d}{{dt}}\left( {a{t^2} + 2bt + c} \right)$

Now again differentiate it we have,

$ \Rightarrow v = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }}\dfrac{d}{{dt}}\left( {a{t^2} + 2bt + c} \right) = \dfrac{1}{{2\sqrt {a{t^2} + 2bt + c} }}\left( {2at + 2b + 0} \right)$
Now simplify this we have,

$ \Rightarrow v = \dfrac{{\left( {at + b} \right)}}{{\sqrt {a{t^2} + 2bt + c} }}$................. (3)
Now from equation (1) we have,

$ \Rightarrow v = \dfrac{{\left( {at + b} \right)}}{x}$..................... (4)

Now acceleration (A) is the rate of change of velocity so differentiate equation (4) w.r.t. t we have,

$ \Rightarrow A = \dfrac{d}{{dt}}v = \dfrac{d}{{dt}}\dfrac{{\left( {at + b} \right)}}{x}$
Now apply divide rule of differentiate $\dfrac{d}{{dt}}\dfrac{u}{v} = \dfrac{{v\dfrac{d}{{dt}}u - u\dfrac{d}{{dt}}v}}{{{v^2}}}$ so use this property in above equation we have,
$ \Rightarrow A = \dfrac{d}{{dt}}\dfrac{{\left( {at + b} \right)}}{x} = \dfrac{{x\dfrac{d}{{dt}}\left( {at + b} \right) - \left( {at + b} \right)\dfrac{d}{{dt}}x}}{{{x^2}}}$
Now differentiate it we have,

$ \Rightarrow A = \dfrac{{x\left( {a + 0} \right) - \left( {at + b} \right)\dfrac{{dx}}{{dt}}}}{{{x^2}}}$

Now from equation (2), $v = \dfrac{{dx}}{{dt}}$ so substitute this value in above equation we have,

$ \Rightarrow A = \dfrac{{ax - \left( {at + b} \right)v}}{{{x^2}}}$
Now from equation (4), $v = \dfrac{{\left( {at + b} \right)}}{x}$ so substitute this value in above equation we have,
$ \Rightarrow A = \dfrac{{ax - \left( {at + b} \right)\dfrac{{\left( {at + b} \right)}}{x}}}{{{x^2}}}$
Now simplify this we have,
$ \Rightarrow A = \dfrac{{a{x^2} - {{\left( {at + b} \right)}^2}}}{{{x^3}}}$
Now from equation (1) we have,
$ \Rightarrow A = \dfrac{{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}}}{{{x^3}}}$

Now simplify this according to the property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,

$ \Rightarrow A = \dfrac{{\left( {{a^2}{t^2} + 2abt + ac} \right) - \left( {{a^2}{t^2} + {b^2} + 2abt} \right)}}{{{x^3}}} = \dfrac{{ac - {b^2}}}{{{x^3}}}$
So as we see that ($ac - {b^2}$) is constant so let, $ac - {b^2}$ = C, where C is some constant.
$ \Rightarrow A = \dfrac{{ac - {b^2}}}{{{x^3}}} = \dfrac{C}{{{x^3}}}$
$ \Rightarrow A = C{x^{ - 3}}$

So acceleration (A) of the particle varies inverse of the cube root of distance x covered by the particle.
So this is the required answer.

Hence option (A) is the required answer.

Note: The trick point here was that the initial distance covered by given in terms of ${x^2}$and not x that is ${x^2} = a{t^2} + 2bt + c$ and thus one must have directly differentiated this to obtain the expression of velocity but however it would have been wrong as we first need to convert this in terms of x, so in general velocity is rate change of position when the position is written as a function of x and t that is $f(x,t)$.