Step-by-step Solution:

We know that velocity is given by: \[ v = \frac{dx}{dt} \] and acceleration is: \[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v. \] Given that: \[ a = v \cdot \frac{dv}{dx} = x - x^2, \] we can write: \[ \int v \, dv = \int (x - x^2) \, dx. \] On integration, we get: \[ \frac{v^2}{2} = \frac{x^2}{2} - \frac{x^3}{3} + c. \] Since the particle is at rest at the origin, \( x = 0 \) and \( v = 0 \), therefore: \[ c = 0. \] The particle comes to rest again when: \[ \frac{x^2}{2} - \frac{x^3}{3} = 0. \] Factoring, we get: \[ x \left( \frac{x}{2} - \frac{x^2}{3} \right) = 0 \implies x = 0 \, \text{or} \, x = \frac{3}{2}. \] Thus, the particle comes to rest again at \( x = \frac{3}{2} \).