In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x² + x. The acceleration of the particle is

The correct option is:
$-\frac{2}{{(2x+1)}^3}$

Step-by-step Derivation:

We are given the relation between time ($t$) and position ($x$) of a moving particle:
$t=x^2+x$

To find the velocity and acceleration, we differentiate this equation with respect to time ($t$). Using the chain rule for differentiation:
$\frac{dt}{dt}=\frac{d}{dt}(x^2+x)$
$1=(2x+1)\cdot \frac{dx}{dt}$

Since velocity ($v$) is defined as the rate of change of position with respect to time ($v=\frac{dx}{dt}$), we can write:
$1=(2x+1)v$
Solving for $v$ gives:
$v=\frac{1}{2x+1}={(2x+1)}^{-1}$

Acceleration ($a$) is the derivative of velocity with respect to time ($a=\frac{dv}{dt}$). Differentiating $v$ with respect to $t$ using the chain rule:
$a=\frac{dv}{dt}=\frac{d}{dt}{(2x+1)}^{-1}$
$a=-1{(2x+1)}^{-2}\cdot \frac{d}{dt}(2x+1)$
$a=-{(2x+1)}^{-2}\cdot \left(2\frac{dx}{dt}\right)$
Substituting $\frac{dx}{dt}=v$:
$a=-2{(2x+1)}^{-2}\cdot v$

Now, we substitute the expression for $v$ back into the acceleration equation:
$a=-2{(2x+1)}^{-2}\cdot {(2x+1)}^{-1}$
$a=-2{(2x+1)}^{-3}$
Simplifying the negative exponent representation:
$a=-\frac{2}{{(2x+1)}^3}$