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The acceleration of a particle starting from rest, varies with time according to the relation a=kt+c. The velocity of the particle after time t will be

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${kt}^{2} + ct$

$\frac{1}{2} {kt}^{2} + ct$

$\frac{1}{2} ({kt}^{2} + ct)$

${kt}^{2} + \frac{1}{2} ct$

answer is B.

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Detailed Solution

$a = kt + c$

$\frac{dv}{dt} = kt + c$

$dv = (kt + c) dt$

$\int dv = \int (kt + c) dt$

$v = \frac{{kt}^{2}}{2} + ct$

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