Q: A particle moves a distance x in time t according to the equation x = (t + 5)–1 . The acceleration of particle is proportional to
(a) (velocity)3/2
(b) (distance)2
(c) (distance)–2
(d) (velocity)2/3
Ans: (a)
Sol: $\displaystyle x = \frac{1}{t+5} $
$ \displaystyle v = \frac{dx}{dt} = \frac{-1}{(t+5)^2} $
Again differentiating w.r.t time
$ \displaystyle a = \frac{dv}{dt} = \frac{2}{(t+5)^3} $
$ \displaystyle a = 2 (\frac{1}{(t+5)^2})^{3/2} $
$\displaystyle a \propto v^{3/2}$