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If the velocity of a particle is $v = A t + B t^{2},$ where A and B are constants, then its acceleration at 1 sec is

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$\frac{3}{2} A + 4 B$

$A + 2 B$

$\frac{3}{2} A + \frac{7}{3} B$

$\frac{A}{2} + \frac{B}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} a = d c / d t \\ a = \frac{d (A t + B t^{2})}{d t} = A + 2 B t \\ a t t = 1 \sec; a = A + 2 B \end{array}$

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