JEE Main 2026 — Motion In One Dimension Question with Solution

Given $x(t)=4t^3-3t$, we find $v(t)=12t^2-3$ and $a(t)=24t$.
Statement A: Setting $x(t)=0$ gives $4t^3-3t=0\Rightarrow t(4t^2-3)=0$, so $t=0$ or $t=\sqrt{3/4}=0.866$. The particle returns to origin at $t=0.866$ units. TRUE.
Statement B: Turning points occur at $v(t)=0\Rightarrow 12t^2-3=0\Rightarrow t=\pm 0.5$. At $t=0.5$: $x=4(0.125)-1.5=-1$; distance = 1 unit. At $t=-0.5$: $x=1$; distance = 1 unit. TRUE.
Statement C: For $t\ge 0$ (particle starts at $t=0$), $a(t)=24t\ge 0$. TRUE.
Statement D: At turning points, distance is 1 unit, not 0.5. FALSE.
Statement E: False because acceleration is non-negative only for $t\ge 0$, and particle does turn back at $t=0.5$. FALSE.
Correct statements: A, B, C Only.