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The displacement of a particle moving in a straight line is given by the expression $x = A t^{3} + B t^{2} + C t + D$ in meters., where is in seconds and A,B,C and D are constants. The ratio between the initial acceleration and initial velocity is

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$\frac{C}{2 B}$

$\frac{2 C}{B}$

$\frac{2 B}{C}$

$2 C$

answer is B.

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Detailed Solution

$x = A t^{3} + B t^{2} + C t + D$

$v = \frac{d x}{d t} = 3 A t^{2} + 2 B t + C; a t t = 0 \Rightarrow V_{0} = C$

$a = \frac{d V}{d t} = 6 A t + 2 B; a t t = 0 \Rightarrow a_{0} = 2 B$

$\Rightarrow \frac{a_{0}}{v_{0}} = \frac{2 B}{c}$

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