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Question

$x^{3}=t^{3}+1 \Rightarrow 3 x^{2} \frac{d x}{d t}=3 t^{2}$
$\mathrm{x}^{2} \mathrm{v}=\mathrm{t}^{2} \Rightarrow 2 \mathrm{x} \frac{\mathrm{dx}}{\ma

Vedclass · Accepted Answer

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

Vedclass · Answer

$x^{3}=t^{3}+1 \Rightarrow 3 x^{2} \frac{d x}{d t}=3 t^{2}$
$\mathrm{x}^{2} \mathrm{v}=\mathrm{t}^{2} \Rightarrow 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{v}+\mathrm{x}^{2} \mathrm{a}=2 \mathrm{t}$
$2 \mathrm{x} \frac{\mathrm{t}^{4}}{\mathrm{x}^{4}}+\mathrm{x}^{2} \mathrm{a}=2 \mathrm{t} \Rightarrow \mathrm{x}^{2} \mathrm{a}=2 \mathrm{t}-\frac{2 \mathrm{t}^{4}}{\mathrm{x}^{3}}$
$a=\frac{2 t\left[x^{3}-t^{3}\right]}{x^{5}} \Rightarrow a=\frac{2 t}{x^{5}}$

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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