The distance x covered by a particle in one dimensional motion varies with time t as `x^(2) = at^(2) +2bt +c` . If the acceleration of the particle depends on x as `x^(-n)` . Where n is an integer , the value of n is

Explanation

To solve the problem, we need to analyze the motion of the particle given the equation \( x^2 = At^2 + 2Bt + C \) and find the integer \( n \) such that the acceleration \( a \) is proportional to \( x^{-n} \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ x^2 = At^2 + 2Bt + C \]

2. **Differentiate with respect to time \( t \)** to find the velocity \( v \):
\[ \frac{d}{dt}(x^2) = \frac{d}{dt}(At^2 + 2Bt + C) \]
Using the chain rule on the left side:
\[ 2x \frac{dx}{dt} = 2At + 2B \]
This simplifies to:
\[ 2xv = 2At + 2B \]
where \( v = \frac{dx}{dt} \).

3. **Rearranging to find \( v \)**:
\[ xv = At + B \quad \text{(Equation 1)} \]

4. **Differentiate Equation 1 with respect to time \( t \)** to find acceleration \( a \):
\[ \frac{d}{dt}(xv) = \frac{d}{dt}(At + B) \]
Using the product rule on the left side:
\[ v \frac{dx}{dt} + x \frac{dv}{dt} = A \]
Substituting \( \frac{dx}{dt} = v \):
\[ v^2 + x \frac{dv}{dt} = A \]
Rearranging gives:
\[ x \frac{dv}{dt} = A - v^2 \]

5. **Express acceleration \( a \)**:
\[ a = \frac{dv}{dt} = \frac{A - v^2}{x} \]

6. **Substituting \( v \) from Equation 1**:
From Equation 1, we have:
\[ v = \frac{At + B}{x} \]
Substituting this into the expression for acceleration:
\[ a = \frac{A - \left(\frac{At + B}{x}\right)^2}{x} \]

7. **Simplifying the expression**:
\[ a = \frac{A - \frac{(At + B)^2}{x^2}}{x} \]
\[ a = \frac{Ax^2 - (At + B)^2}{x^3} \]

8. **Identifying the relationship with \( x^{-n} \)**:
The expression for acceleration can be rewritten as:
\[ a \propto \frac{1}{x^3} \]
This indicates that:
\[ a \propto x^{-3} \]

9. **Conclusion**:
Comparing \( a \propto x^{-n} \) with \( a \propto x^{-3} \), we find that:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).