The relation between time t and distance x fo | vedclass

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Question

$t =m x^{2}+n x$

$\frac{1}{v}=\frac{d t}{d x}=2 m x+n$

$v=\frac{1}{2 m x+n}$

$\frac{d v}{d t}=-\frac{2 m}{(2 m x+n)^{2}}\left(\frac{d x}{d t}

Vedclass · Accepted Answer

$2 {mv}^{3}$

Vedclass · Answer

$t =m x^{2}+n x$

$\frac{1}{v}=\frac{d t}{d x}=2 m x+n$

$v=\frac{1}{2 m x+n}$

$\frac{d v}{d t}=-\frac{2 m}{(2 m x+n)^{2}}\left(\frac{d x}{d t}\right)$

$a=-(2 m) v^{3}$

The relation between time $t$ and distance $x$ for a moving body is given as $t=m x^{2}+n x$, where ${m}$ and ${n}$ are constants. The retardation of the motion is -

(Where $v$ stands for velocity)

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