Question:
The relation between time $\mathrm{t}$ and distance $\mathrm{x}$ for a moving body is given as $\mathrm{t}=\mathrm{mx}^{2}+\mathrm{nx}$, where $\mathrm{m}$ and $\mathrm{n}$ are constants. The retardation of the motion is : (When $v$ stands for velocity)
Correct Option: 1
Solution:
$\mathrm{t}=\mathrm{mx} \mathrm{x}^{2}+\mathrm{nx}$
$\frac{1}{\mathrm{~V}}=\frac{\mathrm{dt}}{\mathrm{dx}}=2 \mathrm{mx}+\mathrm{n}$
$\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}$
$\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{2 \mathrm{~m}}{(2 \mathrm{mx}+\mathrm{n})^{2}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)$
$a=-(2 m) v^{3}$