Physics · 2021
JEE Main · 25 July 2021, Shift 2 · Q4
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:…
The relation between time $\displaystyle t$ and distance $\displaystyle x$ for a moving body is given as $\displaystyle t=m x^2+n x$, where $\displaystyle m$ and $\displaystyle n$ are constants. The retardation of the motion is : (Where $\displaystyle v$ stands for velocity)
Official answer
(1)
$\displaystyle 2 m v^3$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.