The relation between time t and distance x is $t = a x^{2} + b x$ where a and b are constants. The acceleration and velocity relation is acceleration $= - k a v^{3}$ . The value of k is

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answer is 2.

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Detailed Solution

$t = a x^{2} + b x$ Differentiating both sides w.r.t t we get

$1 = (2 a x + b) \frac{d x}{d t} \Rightarrow V = \frac{d x}{d t} = \frac{1}{(2 a x + b)} = {(2 a x + b)}^{- 1}$

$a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t} = \frac{d v}{d x} . v = - {(2 a x + b)}^{- 2} (2 a) v$ $acceleration = - v^{2} 2 av = - 2 {av}^{3}$

k-=2

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