The relation between time $t$ and distance $x$ is $t = a x^{2} + b x,$ where $a$ and $b$ are constants. 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 $- 2 a v^{3}$
Given, $t = a x^{2} + b x$
$\frac{d t}{d t} = 1 = 2 a x \frac{d x}{d t} + b \frac{d x}{d t}$
$v = \frac{d x}{d t} = (\frac{1}{2 a x + b})$
$a = v \frac{d v}{d x} = \frac{1}{2 a x + b} (\frac{- 2 a}{(2 a x + b)^{2}})$
$a = - 2 a v^{3}$

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