Help is greatly appreciated (1 Viewer)

vivekjain0908 · Jun 11, 2022

Dk wut to do in this question

ExtremelyBoredUser · Jun 11, 2022

vivekjain0908 said:
View attachment 35791
Dk wut to do in this question

$a = -v(1+v^2)$
$v\frac{dv}{dx} = -v(1+v^2)$
$\frac{dv}{dx} = -(1+v^2)$
$- \int_{20}^{v} \frac{1}{1+v^2} \, dv = \int_{0}^{x} \, dx$
$\tan^{-1}(20) - \tan^{-1}(v)= x$
$\tan(x) = \tan{\left(\tan^{-1}(20) - \tan^{-1}(v)\right)}$
Let $\alpha = \tan^{-1}(20)$ and $\beta = \tan^{-1}(v)$
$\tan(x) = \tan{\left(\alpha- \beta \right)}$
$\tan(x) = \frac{\tan(\alpha) - tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$
$\tan(x) = \frac{20-v}{1+20v}$
$x = \tan^{-1}(\frac{20-v}{1+20v})$

Help is greatly appreciated (1 Viewer)

vivekjain0908

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ExtremelyBoredUser

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