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The displacement x of a particle at the instant when its velocity is v is given by $ v = \sqrt {3x+16}$ . Its acceleration and initial velocity are

$\begin{array}{1 1} 1.5\;units,4\;units \\ 3\;units,4\;units \\16\;units,1.6\;units \\16 \;units, 3\;units \end{array} $

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Answer : $1.5\;units,4\;units$
$v= \sqrt {3x+16}$
or $v^2 =3x+16$ or $v^2-16=3x$
Comparing with $v^2-u^2=2as$
We get, $u=4 \;units$
$2a= 3$ or $a=1.5\;units$
answered Aug 19, 2014 by meena.p

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