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If the velocity of a particle moving along a straight line is directly proportional to the square of its distance from a fixed point on the line. Then its acceleration is proportional to

\[\begin{array}{1 1}(1)S&(2)S^{2}\\(3)S^{3}&(4)S^{4}\end{array}\]

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Let $S$ be the distance of the particle from a fixed o the lie at time 't' af 'v' be its velocity the $\large\frac{ds}{dt}$
Given $ v \alpha s^2$
ie $v= ks^2$
$\large\frac{dv}{dt} =k.$$ 2s. \large\frac{ds}{dt}$
$\qquad= 2ks.V$
$\qquad= 2ks .Vs$
Acceleration $=2k^2s^3$
Hence 3 is the correct answer.
answered May 19, 2014 by meena.p

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