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If $x = {αt}^{2} - {βt}^{3}$ then velocity of particle when acceleration is 0 is

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$\frac{2 α^{2}}{3 β}$

$\frac{α^{2}}{β}$

$\frac{α^{2}}{3 β}$

$\frac{2 α^{2}}{β}$

answer is D.

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Detailed Solution

$v = 2 αt - 3 {βt}^{2}$

$a = 2 α - 6 βt = 0 \Rightarrow t = \frac{α}{3 β}$
velocity at $t = \frac{α}{3 β}$

$v = 2 α (\frac{α}{3 β}) - 3 {(\frac{α}{3 β})}^{2} = \frac{α^{2}}{3 β}$

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