JEE Main 2019 — Physics solution | Fundamenthol

JEE Main2019Physics

Q.

The position of a particle as a function of time t, is given by x(t) = at + bt²– ct³where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

A
$a + \frac{b ^{2}}{4 c}$
B
$a + \frac{b ^{2}}{3 c}$
C
$a + \frac{b ^{2}}{2 c}$
D
$a + \frac{b ^{2}}{c}$

Solution

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →