A particle moves a distance $\mathrm{x}$ in time $\mathrm{t}$ according to equation $x=(t+5)^{-1}$. The…
A particle moves a distance $\mathrm{x}$ in time $\mathrm{t}$ according to equation $x=(t+5)^{-1}$. The acceleration of particle is proportional to
(velocity) ${ }^{3 / 2}$
$(\text { distance })^2$
(distance $)^{-2}$
(velocity) ${ }^{2 / 3}$
Solution
Given, distance $x=(t+5)^{-1}$
Differentiating Eq. (i) w.r.t. $t$, we get
$\frac{\mathrm{dx}}{\mathrm{dt}}=(\mathrm{v})=\frac{-1}{(\mathrm{t}+5)^2}$
Again, differentiating Eq. (i) w.r.t. t, we get
$\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=(\mathrm{a})=\frac{2}{(\mathrm{t}+5)^3}$
Comparing Eqs. (ii) and (iii), we get
(a) $\propto(v)^{3 / 2}$