A particle starts moving from time t = 0 and its coordinate is given as $$x(t)=4t^{3}-3t$$
A. The particle returns to its original position (origin) 0.866 units later
B. The particle is 1 unit away from the origin at its turning point
C. Acceleration of the particle is non-negative
D. The particle is 0.5 units away from the origin at its turning point
E. Particle never turns back as acceleration is non-negative
Choose the correct answer from the options given below :

A particle starts at $$t = 0$$ with position $$x(t) = 4t^3 - 3t$$. Determine which statements are correct.

$$v(t) = x'(t) = 12t^2 - 3$$

$$a(t) = x''(t) = 24t$$

$$12t^2 - 3 = 0 \implies t^2 = 1/4 \implies t = 1/2$$ (taking $$t > 0$$)

Statement A: "The particle returns to the origin 0.866 units later."

$$x(t) = 0$$: $$4t^3 - 3t = 0 \implies t(4t^2 - 3) = 0 \implies t = 0$$ or $$t = \sqrt{3}/2 \approx 0.866$$.

The particle returns to origin at $$t = \sqrt{3}/2 \approx 0.866$$. TRUE.

Statement B: "The particle is 1 unit away from origin at its turning point."

$$x(1/2) = 4(1/8) - 3(1/2) = 1/2 - 3/2 = -1$$. Distance from origin = 1. TRUE.

Statement C: "Acceleration is non-negative."

$$a(t) = 24t \geq 0$$ for $$t \geq 0$$. TRUE.

Statement D: "The particle is 0.5 units away at its turning point."

The distance is 1, not 0.5. FALSE.

Statement E: "Particle never turns back as acceleration is non-negative."

Despite non-negative acceleration, the initial velocity $$v(0) = -3 < 0$$, so the particle moves in the negative direction first, then turns at $$t = 1/2$$. FALSE.

Correct statements: A, B, C.

The correct answer is Option 2: A, B, C Only.