Q61 / 90 · JEE Main 2021 (25 Jul Shift 2) · Physics

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: (When $v$ stands for velocity)

Solution

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

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

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

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

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

Hence, the correct answer is option: 'A'.