A particle is projected with velocity v_0 alo | vedclass

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Question

$accelaration=v \frac{d v}{d x}$

$	herefore a=\frac{d v}{d t}=\frac{d v}{d t} 	imes \frac{d x}{d x}$

$\left.=\frac{d u}{d x} 	imes \frac{d x}

Vedclass · Accepted Answer

${\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{1/3}}$

Vedclass · Answer

$accelaration=v \frac{d v}{d x}$

$	herefore a=\frac{d v}{d t}=\frac{d v}{d t} 	imes \frac{d x}{d x}$

$\left.=\frac{d u}{d x} 	imes \frac{d x}{d t} 	ext { (but } \frac{d x}{d t}=vight)$

$\Rightarrow a =\frac{v d v}{d x}$

$\Rightarrow \quad \frac{V d v}{d x}=a x^{2}$

$\Rightarrow \quad v d v=\left(\alpha x^{2}ight) d x$

$\Rightarrow \quad \int v d v=\int \alpha x^{2} d x$

$\Rightarrow \quad \frac{v^{2}}{2}=. \alpha \frac{x^{3}}{3}$

$\Rightarrow \quad x^{3}=\frac{3 v^{2}}{2 \alpha}$

$\Rightarrow \quad x=\left(\frac{3 v^{2}}{2 \alpha}ight)^{\frac{1}{3}}$

A particle is projected with velocity $v_0$ along $x-$ axis and its decelaration on the particle is a $a= -\alpha x^2$. The distance at which the particle stops is

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