Q 11-02-125JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
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 zero acceleration, then its velocity will be
Answer: (A) $a + \dfrac{b^2}{3c}$
$$v = a + 2bt - 3ct^2,\qquad \frac{dv}{dt} = 2b - 6ct = 0 \Rightarrow t = \frac{b}{3c}$$
$$v = a + \frac{2b^2}{3c} - \frac{3cb^2}{9c^2} = a + \frac{b^2}{3c}$$
Solution by Sreeraj P, M.Sc Physics