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

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$- \frac{1}{4x^3}$

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(B) Given the position-time relation: $x^2 = 2 + t$.Differentiating both sides with respect to $t$:$2x \frac{dx}{dt} = 1$.Since velocity $v = \frac{dx}{dt}$,we have $2xv = 1$,which implies $v = \frac{1}{2x}$.Acceleration $a$ is given by $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v = \frac{1}{2} x^{-1}$ with respect to $x$:$\frac{dv}{dx} = \frac{1}{2} (-1) x^{-2} = -\frac{1}{2x^2}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (\frac{1}{2x}) 	imes (-\frac{1}{2x^2}) = -\frac{1}{4x^3}$.

$A$ particle moves in a straight line and its position $x$ at time $t$ is given by $x^2 = 2 + t$. Its acceleration is given by

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