A particle moves a distance $$x$$ in time $$t$$ according to equation $$x = {\left( {t + 5} \right)^{ - 1}}.$$   The acceleration of particle is proportional to

Solution :
$$\eqalign{ & x = \frac{1}{{t + 5}} \cr & \therefore v = \frac{{dx}}{{dt}} = \frac{{ - 1}}{{{{\left( {t + 5} \right)}^2}}} \cr & \therefore a = \frac{{{d^2}x}}{{d{t^2}}} = \frac{2}{{{{\left( {t + 5} \right)}^3}}} = 2{x^3} \cr & {\text{Now}}\,\frac{1}{{\left( {t + 5} \right)}} \propto {v^{\frac{1}{2}}} \cr & \therefore \frac{1}{{{{\left( {t + 5} \right)}^3}}} \propto {v^{\frac{3}{2}}} \propto a \cr} $$