A particle starts moving from time t=0 and its coordinate is given...

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 $t \geq 0$ (particle starts at $t=0$), $a(t) = 24t \geq 0$. TRUE.
Statement D: At turning points, distance is 1 unit, not 0.5. FALSE.
Statement E: False because acceleration is non-negative only for $t \geq 0$, and particle does turn back at $t = 0.5$. FALSE.
Correct statements: A, B, C Only.