Question

$\huge\text{Question:-}$


The velocity $v$ of a parachute falling vertically satisfies the equation $v \ \ \dfrac{dv}{dx} = g ( 1 - \dfrac{v²}{k²} )$, where $g$ and $k$ are constants. If both $v$ and $x$ are zero initially, find $v$ in terms of $x$.


@DIFFRENTIAL EQUATIONS
#Class - 12​

Answer 1

Answer:

Given:-

$= v \frac{dv}{dx} = g \left(1 - \frac{ {v}^{2} }{ {k}^{2} } \right) \] \: \text{or} \: v \: dv = g \: \left( \frac{ {k}^{2} - {v}^{2} }{ {k}^{2} } \right) \] \: dx$

ㅤㅤㅤㅤㅤOR

$\frac{v \: dv}{ {k}^{2} - {v}^{2} } = \frac{g}{ {k}^{2} } \: dx$

$\text{Multiplying both sides by -2 and integrating}$

we obtain

[tex]⇒\displaystyle \int\limits_{}^{} \frac{ - 2v \: dv}{ {k}^{2} - {v}^{2} } =

\displaystyle \int\limits_{}^{} - \frac{2g}{ {k}^{2} } dx + c[/tex]

$⇒kg( {k}^{2} - {v}^{2} ) = - \frac{2g}{ {k}^{2} } x + c$

$\text{when \: } x = 0 ,v = 0 ,\text{therefore ,} log \: {k}^{2} = 0 + c \: \: \: ..(1)$

$(\text{we \: assume \: that \: } {k}^{2} - {v}^{2} > 0)$

$\text{Subtracting \: (2) \: from \: (1)}$

we get,

$⇒ log( {k}^{2} - {v}^{2} ) - log \: {k}^{2} = - \frac{2gx}{ {k}^{2} }$

$⇒ log \left( \frac{ {k}^{2} - {v}^{2} }{ {k}^{2} } \right) \] = - \frac{2gx}{ {k}^{2} }$

$⇒1 - \frac{ {v}^{2} }{ {k}^{2} } = e \: - \frac{ 2gx}{ {k}^{2} }$

$⇒ {v}^{2} = {k}^{2} (1 - {e}^{ - 2gx/ {k}^{2} }$

$⇒v = k \sqrt{1 - {e}^{ - 2gx/ {k}^{2} } }$

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Answer 2

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