Q: The distance x covered by a particle in one dimensional motion varies with time t as x^{2} = at^{2} + 2 bt + c . If the acceleration of particle depends on x as x^{-n} , where n is an integer , the value of n is …….

Sol: x^{2} = at^{2} + 2 bt + c

$ \large x = (at^2 + 2 bt + c)^{1/2} $

Differentiating with respect to time t ,

$ \large \frac{dx}{dt} = \frac{1}{2}(at^2 + 2 bt + c)^{-1/2}(2at + 2b) $

Acceleration , $ \displaystyle A = \frac{d^2x}{dt^2} = \frac{1}{2}[ (at^2 + 2 bt + c)^{-1/2}(2a) + (2at + 2b)(-\frac{1}{2})(at^2 + 2bt + c)^{-3/2}(2at + 2b)] $

$\large Ax = a – (\frac{at+b}{x})^2 $

$\large Ax = \frac{ax^2 -(at+b)^2 }{x^2} $

$\large Ax = \frac{a(at^2 + 2bt + c)-(at+b)^2}{x^2}$

$\large A = \frac{ac-b^2}{x^3} $

$\large A \propto \frac{1}{x^3} $

$\large A \propto x^{-3} $

Hence , n = 3