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(B) Given the displacement equation: $x^3 = t^3 + 1$. Differentiating both sides with respect to time $t$: $3x^2 \frac{dx}{dt} = 3t^2$ $x^2 v = t^2$,w

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$\frac{2t}{x^5}$

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(B) Given the displacement equation: $x^3 = t^3 + 1$. Differentiating both sides with respect to time $t$: $3x^2 \frac{dx}{dt} = 3t^2$ $x^2 v = t^2$,where $v = \frac{dx}{dt}$ is the velocity. $v = \frac{t^2}{x^2}$. Differentiating $x^2 v = t^2$ with respect to time $t$ using the product rule: $2x \frac{dx}{dt} v + x^2 \frac{dv}{dt} = 2t$ Since $a = \frac{dv}{dt}$,we have: $2x v^2 + x^2 a = 2t$ Substitute $v = \frac{t^2}{x^2}$: $2x (\frac{t^2}{x^2})^2 + x^2 a = 2t$ $2x \frac{t^4}{x^4} + x^2 a = 2t$ $x^2 a = 2t - \frac{2t^4}{x^3}$ Since $x^3 = t^3 + 1$,we have $t^3 = x^3 - 1$: $x^2 a = 2t - \frac{2t(t^3)}{x^3} = 2t - \frac{2t(x^3 - 1)}{x^3} = 2t - 2t + \frac{2t}{x^3} = \frac{2t}{x^3}$ $a = \frac{2t}{x^3 \cdot x^2} = \frac{2t}{x^5}$.

$A$ point moves such that its displacement as a function of time is given by $x^3 = t^3 + 1$. Its acceleration as a function of time $t$ will be

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