The displacement ' $x$ ' and time of travel 't' for a particle moving an a straight line are related as $t=$ $\sqrt{(x+1)(x-1)}$. Its acceleration at a time $t$ is
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The displacement ' $x$ ' and time of travel 't' for a particle moving an a straight line are related as $t=$ $\sqrt{(x+1)(x-1)}$. Its acceleration at a time $t$ is

(A) $\frac{1}{x}-\frac{1}{x^{2}}$

(B) $\frac{1}{x^{3}}$

(C) $\frac{-1^{2}}{x^{3}}$

(D) $\frac{-1}{x^{2}}$
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Best Answer

Solution —

$t_{2}=x_{2}-1$

$x_{2}=1+t_{2}$

$x \frac{d x}{d t}=t$

$\Rightarrow \quad\left(\frac{d x}{d t}\right)^{2}+x \frac{d^{2} x}{d t^{2}}=1$

$\Rightarrow \quad \frac{x d^{2} x}{d t^{2}}=1-\left(\frac{d x}{d t}\right)^{2}=1-\left(\frac{t}{x}\right)^{2}$

$=\frac{x^{2}-t^{2}}{x^{2}}=\frac{1}{x^{2}} \Rightarrow a=\frac{d^{2} x}{d t^{2}}=\frac{1}{x^{3}}$

So, The correct option of this question is (B).

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