JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Kinematics

For $x(t)=4t^3-3t$: setting $x=0$ gives $t(4t^2-3)=0$, so apart from $t=0$ the particle returns to the origin at $t=\frac{\sqrt{3}}{2}\approx 0.866$ units of time, confirming statement A. The velocity is $v=\frac{dx}{dt}=12t^2-3$, which vanishes at $t=\frac{1}{2}$; there $x=4\left(\frac{1}{8}\right)-3\left(\frac{1}{2}\right)=-1$, so the turning point is $1$ unit from the origin, confirming statement B and making statement D false. The acceleration is $a=\frac{dv}{dt}=24t\ge 0$ for $t\ge 0$, confirming statement C. Statement E is false, because the velocity does vanish at $t=\frac{1}{2}$ so the particle does turn back even though the acceleration stays non-negative. The correct statements are A, B and C.