A particle moves a distance x in time t according to the equation x = (t + 5)^–1 . The acceleration

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

$\displaystyle a = \frac{dv}{dt} = \frac{2}{(t+5)^3}$

$\displaystyle a = 2 (\frac{1}{(t+5)^2})^{3/2}$

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