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The position x of a particle varies with time as $x=at^2-bt^3$. The acceleration of the particle will be zero at time t equal to

\[(a)\;\frac{a}{b}\quad (b)\;\frac{2a}{3b} \quad (c)\;\frac{a}{3b} \quad (d)\;zero\]

1 Answer

Hence c is the correct answer


answered Jun 21, 2013 by meena.p
edited Jan 24, 2014 by meena.p

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