A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O given by `F=-(k)/(x^(2))hat(i)`. If the particle starts from rest at x = a. The speed of it will attain to reach at distance x from the origin O will be

Options

A. `sqrt((2k)/(m))[(a-x)/(ax)]^(1//2)`
B. `sqrt((2k)/(m))[(a+x)/(ax)]^(1//2)`
C. `sqrt(k/m)[(ax)/(a-x)]`
D. `sqrt((m)/(2k))[(a-x)/(ax)]^(1//2)`

Explanation

To solve the problem, we need to find the speed \( v \) of a particle of mass \( m \) moving under the influence of a force \( F = -\frac{k}{x^2} \hat{i} \), starting from rest at \( x = a \) and reaching a distance \( x \) from the origin \( O \).

### Step-by-Step Solution:

1. **Identify the Force and Relation to Acceleration**:
The force acting on the particle is given by:
\[ F = -\frac{k}{x^2} \]
According to Newton's second law, we can relate force to acceleration:
\[ F = m \cdot a \]
where \( a = \frac{dv}{dt} \).

2. **Express Acceleration in Terms of Velocity**:
We can express acceleration \( a \) as:
\[ a = v \frac{dv}{dx} \]
Thus, we can rewrite the force equation as:
\[ m \cdot v \frac{dv}{dx} = -\frac{k}{x^2} \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ v \, dv = -\frac{k}{m} \frac{1}{x^2} dx \]

4. **Integrating Both Sides**:
We will integrate both sides. The left side integrates from \( 0 \) to \( v \) (since the particle starts from rest) and the right side integrates from \( a \) to \( x \):
\[ \int_0^v v \, dv = -\frac{k}{m} \int_a^x \frac{1}{x^2} \, dx \]

5. **Calculating the Integrals**:
The left side becomes:
\[ \frac{v^2}{2} \]
The right side can be integrated as follows:
\[ -\frac{k}{m} \left[-\frac{1}{x}\right]_a^x = -\frac{k}{m} \left(-\frac{1}{x} + \frac{1}{a}\right) = \frac{k}{m} \left(\frac{1}{x} - \frac{1}{a}\right) \]

6. **Setting the Integrals Equal**:
Equating both sides gives:
\[ \frac{v^2}{2} = \frac{k}{m} \left(\frac{1}{x} - \frac{1}{a}\right) \]

7. **Solving for \( v^2 \)**:
Multiplying both sides by 2:
\[ v^2 = \frac{2k}{m} \left(\frac{1}{x} - \frac{1}{a}\right) \]

8. **Finding \( v \)**:
Taking the square root to find \( v \):
\[ v = \sqrt{\frac{2k}{m} \left(\frac{1}{x} - \frac{1}{a}\right)} \]

### Final Result:
The speed \( v \) of the particle when it reaches a distance \( x \) from the origin is:
\[ v = \sqrt{\frac{2k}{m} \left(\frac{1}{x} - \frac{1}{a}\right)} \]