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))/(2))`
B. `((v_(0)^(2))/(3))^(1//3)`
C. `((2v_(0)^(2))/(3))^(1//3)`
D. `((3v_(0)^(2))/(2))^(1//3)`

Answer

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

### Step-by-Step Solution:

1. **Understand the relationship between acceleration, velocity, and position:**
The acceleration \( a \) can be expressed in terms of velocity \( v \) and position \( x \) using the chain rule:
\[ a = v \frac{dv}{dx} \]
Here, \( a \) is also given as \( -x^2 \).

2. **Set up the equation:**
Equating the two expressions for acceleration, we have:
\[ -x^2 = v \frac{dv}{dx} \]

3. **Rearranging the equation:**
We can rearrange this to separate variables:
\[ -x^2 dx = v dv \]

4. **Integrate both sides:**
We will integrate both sides. The left side will be integrated with respect to \( x \) from \( 0 \) to \( s \) (the distance where the particle stops), and the right side will be integrated with respect to \( v \) from \( v_0 \) to \( 0 \):
\[ \int_{0}^{s} -x^2 \, dx = \int_{v_0}^{0} v \, dv \]

5. **Calculating the integrals:**
- The left integral:
\[ \int -x^2 \, dx = -\frac{x^3}{3} \Big|_{0}^{s} = -\frac{s^3}{3} \]
- The right integral:
\[ \int v \, dv = \frac{v^2}{2} \Big|_{v_0}^{0} = \frac{0^2}{2} - \frac{v_0^2}{2} = -\frac{v_0^2}{2} \]

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

7. **Solving for \( s^3 \):**
Removing the negative signs and rearranging gives:
\[ \frac{s^3}{3} = \frac{v_0^2}{2} \]
Multiplying both sides by 3:
\[ s^3 = \frac{3v_0^2}{2} \]

8. **Finding \( s \):**
Taking the cube root of both sides, we find:
\[ s = \left(\frac{3v_0^2}{2}\right)^{1/3} \]

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