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The displacement x of a particle at time t moving along a straight line path is given by $x^{2} = {at}^{2} + 2 bt + c$ Where a, b and c are constants the acceleration of the particle varies as

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$x^{- 1}$

$x^{- 2}$

$x^{- 3}$

$x^{- 4}$

answer is C.

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Detailed Solution

$x^{2} = {at}^{2} + 2 bt + c$

Diff wrt ‘t’ we get xv = at + b

Again diff wrt ‘t’ we get $x \frac{dv}{dt} = a - v^{2}$

Acceleration $= \frac{a - v^{2}}{x}$

$A = \frac{a - {(\frac{at + b}{x})}^{2}}{x}$

Solving $A α x^{- 3}$

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