Relation between time t and distance x for a moving body is given as t=m x^{2}+n x , where {m} and {

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2.Motion in Straight Line

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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)

A

$2 n^{2} v^{3}$

B

$2 {mv}^{3}$

C

$2 n v^{3}$

D

$2 {mnv}^{3}$

(JEE MAIN-2021)

Solution

$t =m x^{2}+n x$

$\frac{1}{v}=\frac{d t}{d x}=2 m x+n$

$v=\frac{1}{2 m x+n}$

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

$a=-(2 m) v^{3}$

Standard 11

Physics

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