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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) $a + \frac{b^2}{2c}$

Show Answer

Answer: (2)


Learning Progress: Step 26 of 32 in this series