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

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

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(B) Given the position equation: $x^2 = t + 2$.Differentiating both sides with respect to time $t$:$2x \frac{dx}{dt} = 1 \Rightarrow \frac{dx}{dt} = \frac{1}{2x}$.Now,differentiate the velocity $v = \frac{dx}{dt} = \frac{1}{2} x^{-1}$ with respect to time $t$ to find acceleration $a$:$a = \frac{dv}{dt} = \frac{d}{dt} (\frac{1}{2} x^{-1}) = \frac{1}{2} (-1) x^{-2} \frac{dx}{dt}$.Substitute $\frac{dx}{dt} = \frac{1}{2x}$ into the equation:$a = -\frac{1}{2x^2} 	imes \frac{1}{2x} = -\frac{1}{4x^3}$.

$A$ 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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