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  4. The displacement x of a particle at the instant when its velocity v is given by v = √3x+16. Its acceleration and initial velocity are

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

JIPMERJIPMER 2010Motion in a Straight Line

Solution:

Given : $v = \sqrt{3x + 16}$
Squaring both sides, we get
$ v^2 = 3x + 16$ or $v^2 - 16 = 3x$
Comparing with $v^2 -u^2 = 2aS,$ we get,
$u = 4$ units
$2a = 3$ or $a = 15$ units