Physics · 2019
JEE Main · 9 April 2019, Shift 2 · Q2
The position of a particle as a function of time t, is given by x( t )= at + bt^2- ct^3 where a, b and c are constants. When the particle attains…
The position of a particle as a function of time $\displaystyle t$, is given by
$$x(\mathrm{t})=\mathrm{at}+\mathrm{bt}^2-\mathrm{ct}^3
$$
where $\displaystyle \mathrm{a}, \mathrm{b}$ and c are constants. When the particle attains zero acceleration, then its velocity will be :
Official answer
(2)
$\displaystyle a+\frac{b^2}{3 c}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.