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2.Motion in Straight Line
hard
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