Question

If the differential equation of a body falling from rest under gravity is given by $v\frac{dv}{dx}+\frac{n^2}{g}{v}^2=g$, then the velocity of the body is given by $v^2=\frac{g^2}{n^2}(1-e^{-2n^{2}x/g})$.

• A. True
• B. False

Answer 1

The correct option is A True

Given differential equation is $v\frac{dv}{dx}+\frac{n^2{v}^2}{g}=g$,

$⟹\frac{dv}{dx}=\frac{g}{v}-\frac{n^{2}v}{g}$Separating the variables and integrating gives,

${\int }_0^v\frac{gv}{g^2-n^2{v}^2}dv={\int }_0^{x}dx$,

$⟹\ln \left(\frac{g^2}{g^2-n^2{v}^2}\right)=\frac{2n^{2}x}{g}$

$⟹v^2=\frac{g^2}{n^2}(1-e^{-\frac{2n^{2}x}{g}})$