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} $

$\displaystyle a \propto v^{3/2}$