Position of a particle as a function of time is given as `x^2 = at^2 + 2bt + c`, where a, b, c are constants. Acceleration of particle varies with `x^(-n)` then value of n is.

Answer

To solve the given problem, we start with the equation for the position of a particle as a function of time:

\[ x^2 = at^2 + 2bt + c \]

where \( a \), \( b \), and \( c \) are constants. We need to find the relationship between the acceleration of the particle and its position \( x \) in the form of \( x^{-n} \).

### Step 1: Differentiate the position equation with respect to time \( t \)

We differentiate both sides of the equation with respect to \( t \):

\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(at^2 + 2bt + c) \]

Using the chain rule on the left side, we have:

\[ 2x \frac{dx}{dt} = 2at + 2b \]

### Step 2: Simplify the equation

We can simplify this equation by dividing both sides by 2:

\[ x \frac{dx}{dt} = at + b \]

### Step 3: Differentiate again to find acceleration

Now, we differentiate again with respect to \( t \):

\[ \frac{d}{dt}(x \frac{dx}{dt}) = \frac{d}{dt}(at + b) \]

Using the product rule on the left side:

\[ \frac{dx}{dt} \frac{dx}{dt} + x \frac{d^2x}{dt^2} = a \]

This can be rewritten as:

\[ \left(\frac{dx}{dt}\right)^2 + x \frac{d^2x}{dt^2} = a \]

### Step 4: Substitute for acceleration

Let \( f = \frac{d^2x}{dt^2} \) (acceleration). Then we can rewrite the equation as:

\[ \left(\frac{dx}{dt}\right)^2 + x f = a \]

### Step 5: Solve for acceleration

Rearranging gives us:

\[ xf = a - \left(\frac{dx}{dt}\right)^2 \]

### Step 6: Express \( \frac{dx}{dt} \) in terms of \( x \)

From the earlier equation \( x \frac{dx}{dt} = at + b \), we can express \( \frac{dx}{dt} \):

\[ \frac{dx}{dt} = \frac{at + b}{x} \]

### Step 7: Substitute back into the acceleration equation

Substituting this expression into the acceleration equation:

\[ f = \frac{a - \left(\frac{at + b}{x}\right)^2}{x} \]

### Step 8: Simplify the expression for acceleration

After substituting and simplifying, we find:

\[ f = \frac{a - \frac{(at + b)^2}{x^2}}{x} \]

This leads to:

\[ f = \frac{ax^2 - (at + b)^2}{x^3} \]

### Step 9: Identify the relationship with \( x^{-n} \)

From the derived expression, we see that:

\[ f \propto \frac{1}{x^3} \]

Thus, we can conclude that the acceleration \( f \) varies with \( x^{-3} \).

### Conclusion

Therefore, the value of \( n \) is:

\[ \boxed{3} \]