A particle is projected with velocity `v_(0)` along x - axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. `a=-x^(2)`. The distance at which particle stops is -

Options

A. ` sqrt((3v_0)/(2alpha) ) `
B. `((3v_0)/(2alpha))^(1//3) `
C. ` sqrt((2v_0^(2))/(3alpha)) `
D. ` ((3v_0^(2))/(2alpha))^(1//3)`

Explanation

To solve the problem, we need to find the distance at which a particle, projected with an initial velocity \( v_0 \) along the x-axis, stops under the influence of a deceleration that is proportional to the square of its distance from the origin. The deceleration is given by the equation \( a = -x^2 \).

### Step-by-Step Solution:

1. **Understanding the Deceleration**:
The deceleration \( a \) is given by:
\[ a = -\alpha x^2 \]
where \( \alpha \) is a proportionality constant.

2. **Relating Acceleration to Velocity**:
We know that acceleration can also be expressed in terms of velocity and displacement:
\[ a = \frac{dV}{dt} = V \frac{dV}{dx} \]
Therefore, we can write:
\[ V \frac{dV}{dx} = -\alpha x^2 \]

3. **Separating Variables**:
Rearranging the equation gives:
\[ V dV = -\alpha x^2 dx \]

4. **Integrating Both Sides**:
We will integrate both sides. The left side will be integrated from \( v_0 \) to \( 0 \) (since the particle stops), and the right side will be integrated from \( 0 \) to \( X \) (the distance at which it stops):
\[ \int_{v_0}^{0} V dV = -\alpha \int_{0}^{X} x^2 dx \]

5. **Calculating the Integrals**:
The left side integral becomes:
\[ \left[ \frac{V^2}{2} \right]_{v_0}^{0} = 0 - \frac{v_0^2}{2} = -\frac{v_0^2}{2} \]
The right side integral becomes:
\[ -\alpha \left[ \frac{x^3}{3} \right]_{0}^{X} = -\alpha \left( \frac{X^3}{3} - 0 \right) = -\frac{\alpha X^3}{3} \]

6. **Setting the Integrals Equal**:
Now we set the results of the integrals equal to each other:
\[ -\frac{v_0^2}{2} = -\frac{\alpha X^3}{3} \]

7. **Solving for \( X \)**:
Rearranging gives:
\[ \frac{v_0^2}{2} = \frac{\alpha X^3}{3} \]
Multiplying both sides by 3:
\[ \frac{3v_0^2}{2} = \alpha X^3 \]
Now, solving for \( X^3 \):
\[ X^3 = \frac{3v_0^2}{2\alpha} \]
Taking the cube root:
\[ X = \left( \frac{3v_0^2}{2\alpha} \right)^{1/3} \]

### Final Answer:
The distance at which the particle stops is:
\[ X = \left( \frac{3v_0^2}{2\alpha} \right)^{1/3} \]