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Q.

The relation between time land distance x is $t = {ax}^{2} + bx$ where a and b are constants. The acceleration and velocity relation is acceleration $= - {kav}^{3}$ . The value of ‘k’ is

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Detailed Solution

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Given

$t = a x^{2} + b x$

Diff wrt ‘t’ on both sides

$1 = a .2 x . \frac{d x}{d t} + b . \frac{d x}{d t}$

$\begin{array}{l} \Rightarrow 2 a v x + b v = 1 \\ \Rightarrow v (2 a x + b) = 1 \end{array}$

$∵ v = \frac{1}{(2 a x + b)}$

but $a = v . \frac{d v}{d x}$

$a = v . \frac{d v}{d x} = v . \frac{d}{d x} (\frac{1}{2 a x + b})$

$= v [\frac{- 2 a}{{(2 a x + b)}^{2}}] = \frac{- 2 a v}{{(2 a x + b)}^{2}} = - 2 a v (v^{2}) = - 2 a v^{3}$

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The relation between time land distance x is t=ax2+bx where a and b are constants. The acceleration and velocity relation is acceleration =−kav3 . The value of ‘k’ is