It is given that $t = p x^{2} + q x$ , where $x$ is displacement and $t$ is time. The acceleration of particle at origin is

- \frac{2 p}{q^{3}}

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- \frac{2 q}{p^{3}}

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\frac{2 p}{q^{3}}

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\frac{2 q}{p^{3}}

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Solution

The correct option is D $- \frac{2 p}{q^{3}}$
$t = p x^{2} + q x$
$Differentiate with respect to time,$
$t h u s, 1 = 2 p x . \frac{d x}{d t} + q . \frac{d x}{d t}$
$o r 2 p x v + q v = 1 (A t x = 0, q v = 1 o r v = \frac{1}{q})$
$Again differentiate with respect to time,$
$H e n c e, 2 p x a + 2 p v^{2} + q a = 0$
$A t x = 0, p u t v = \frac{1}{q}, t h u s w e g e t a = \frac{- 2 p}{q^{3}}$

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