Given the relationship between time $ t $ and position $ x $ of a moving particle:

$ t = x^2 + x $

First, find the derivative of $ t $ with respect to $ x $:

$ \frac{d t}{d x} = 2x + 1 $

The velocity $ v $ is the inverse of this derivative, as it's given by the derivative of position $ x $ with respect to time $ t $:

$ v = \frac{d x}{d t} = \frac{1}{2x + 1} $

Next, calculate the derivative of $ v $ with respect to $ x $:

$ \frac{d v}{d x} = \frac{-2}{(2x + 1)^2} $

The acceleration $ a $ is then the product of velocity $ v $ and the derivative of velocity with respect to $ x $:

$ a = v \cdot \frac{d v}{d x} = \frac{1}{2x + 1} \cdot \left(\frac{-2}{(2x + 1)^2}\right) $

Simplifying this expression, we find:

$ a = -\frac{2}{(2x + 1)^3} $