The relation between time $t$ and distance of $x$ s is $t = a x^{2} + b x$ where $a$ and $b$ are constant. The acceleration is

- 2 a b v^{2}

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2 b v^{3}

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- 2 a v^{3}

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2 a v^{2}

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Solution

The correct option is C $- 2 a v^{3}$
$t = a x^{2} + b x \Rightarrow 1 = 2 a x (\frac{d x}{d t}) + b (\frac{d x}{d t}) \Rightarrow 1 = 2 a x v + b v \Rightarrow v = \frac{1}{2 a x + b} N o w, a = v \frac{d v}{d x} \Rightarrow v \frac{- 2 a}{{(2 a x + b)}^{2}} = - 2 a v^{3}$

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The correct option is C $- 2 a v^{3}$
$t = a x^{2} + b x \Rightarrow 1 = 2 a x (\frac{d x}{d t}) + b (\frac{d x}{d t}) \Rightarrow 1 = 2 a x v + b v \Rightarrow v = \frac{1}{2 a x + b} N o w, a = v \frac{d v}{d x} \Rightarrow v \frac{- 2 a}{{(2 a x + b)}^{2}} = - 2 a v^{3}$