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The acceleration of a particle moving in a straight line varies with its displacement as $a=2s$ velocity of particle is zero at zero displacement. The corresponding velocity displacement equation is

\[(a)\;v=\sqrt {2s} \quad (b)\;v=\pm \sqrt 2 s \quad (c)\;v^2=\frac{2}{s} \quad (d)\;v=\pm \frac{2}{s}\]

1 Answer

$v \large\frac{dv}{ds}$$=2s$
$\int_{0}^{v} vdv=\int_{0}^{s} 2sds$
$v=\pm \sqrt 2 s$
Hence b is the correct answer.
answered Jul 2, 2013 by meena.p
edited May 24, 2014 by lmohan717

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