Question

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

• A. (velocity)$^{3/2}$
• B. (distance)$^2$
• C. (distance)$^{-2}$
• D. (velocity)$^{2/3}$

Answer 1

The Correct Option is A

Distance, $x=(t+5)^{-1}$ $20\,mm $ $\quad\quad$...(i) Velocity, $v=\frac{dx}{dt}$ $=\frac{d}{dt}(t+5)^{-1}$ $15\,mm = - (t+5)^{-2}$ $20\,mm $ $\quad\quad$ ...(ii) Acceleration, $a=\frac{dv}{dt}$ =$\frac{d}{dt}[-(t+5)^{-2}]$ $25\,mm$ $= 2(t+5)^{-3}$ $20\,mm $ $\quad\quad$...(iii) From equation (ii), we get $20\,mm$ $v^{3/2}$= $(t+5)^{-3}$ $2\quad\quad x^3 = (t+5)^{-3}$ Substituting this in equation (iii), we get Acceleration, $a = 2x^3$ or a $\propto$ (distance)$^{3}$ Hence option ($(velocity)^{3/2}$) is correct.