JEE PYQ: Motion In One Dimension Question 26
Question 26 - 2019 (09 Apr Shift 2)
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:
(1) $a + \frac{b^2}{4c}$
(2) $a + \frac{b^2}{3c}$
(3) $a + \frac{b^2}{c}$
(4) $a + \frac{b^2}{2c}$