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Question

(b)

$x^2=t+2 \Rightarrow \frac{1}{x^2}=\frac{1}{t+2}$

$\Rightarrow x=\sqrt{t+2}$

$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{\frac{1}{2}-1

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)

$x^2=t+2 \Rightarrow \frac{1}{x^2}=\frac{1}{t+2}$

$\Rightarrow x=\sqrt{t+2}$

$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{\frac{1}{2}-1}$

$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{-\frac{1}{2}}$

$\Rightarrow \frac{d^2 x}{d t^2}=\frac{1}{2}\left(-\frac{1}{2}ight)^{(i)}(t+2)^{-\frac{1}{2}-1}$

$\Rightarrow a=-\frac{1}{4}(t+2)^{-\frac{3}{2}}=-\frac{1}{4(t+2)} 	imes \frac{1}{(t+2)^{\frac{1}{2}}}=-\frac{1}{4} 	imes \frac{1}{x^2} 	imes \frac{1}{x}$

$\Rightarrow a=-\frac{1}{4 x^3}$

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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