In some appropriate units,the time (t) and po | vedclass

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(B) Given the relation: $t = x^2 + x$. Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2x + 1$. Since velocity $v = \frac{dx}{dt}$,we

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

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(B) Given the relation: $t = x^2 + x$. Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2x + 1$. Since velocity $v = \frac{dx}{dt}$,we have $\frac{1}{v} = 2x + 1$,which implies $v = (2x + 1)^{-1}$. Acceleration $a$ is given by $a = v \frac{dv}{dx}$. Differentiating $v$ with respect to $x$: $\frac{dv}{dx} = -1(2x + 1)^{-2} 	imes 2 = -2(2x + 1)^{-2}$. Substituting $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (2x + 1)^{-1} 	imes [-2(2x + 1)^{-2}]$. Therefore,$a = -2(2x + 1)^{-3} = -\frac{2}{(2x + 1)^3}$.

In some appropriate units,the 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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