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# The relation between

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)

1. $2 \mathrm{mv}^{3}$

2. $2 \mathrm{mnv}^{3}$

3. $2 \mathrm{nv}^{3}$

4. $2 n^{2} v^{3}$

Correct Option: 1

Solution:

$\mathrm{t}=\mathrm{mx} \mathrm{x}^{2}+\mathrm{nx}$

$\frac{1}{\mathrm{~V}}=\frac{\mathrm{dt}}{\mathrm{dx}}=2 \mathrm{mx}+\mathrm{n}$

$\mathrm{v}=\frac{1}{2 \mathrm{mx}+\mathrm{n}}$

$\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{2 \mathrm{~m}}{(2 \mathrm{mx}+\mathrm{n})^{2}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)$

$a=-(2 m) v^{3}$