In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $

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In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t=x^2+x$. The acceleration of the particle is

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(d) : Given, $t=x^2+x$ Differentiating with respect to ' $t$ ' both sides, $ \begin{aligned} \Rightarrow \quad \frac{d t}{d t} & =2 x \frac{d x}{d t}+\frac{d x}{d t} \ \Rightarrow \quad 1 & =2 x v+v \ \Rightarrow \quad 1 & =v(2 x+1) \ v & =\frac{1}{2 x+1} \end{aligned} $ Again differentiating eq (i) with respect to ' $t$ ', $ \begin{aligned} & 0=2\left(\frac{d x}{d t} \cdot v+x \frac{d v}{d t}ight)+\frac{d v}{d t} \ & 0=2\left(v^2+x \cdot aight)+a \ & 0=2 v^2+2 x a+a \ & 0=\left[2 v^2+a(2 x+1)ight] \ & \Rightarrow 2 v^2=-a(2 x+1) \end{aligned} $ Using equation (ii), we get $ \begin{aligned} & 2\left(\frac{1}{2 x+1}ight)^2=-a(2 x+1) \ & 	herefore a=\frac{-2}{(2 x+1)^3} \end{aligned} $