Motion in a Straight Line Part 2

Concept:The motion is analysed using the position function and its derivatives (velocity and acceleration).Explanation:At $t = 0$, $x(0) = 0$.Return to origin means $x(t) = 0$.$4t^3 - 3t = 0 \Rightarrow t(4t^2 - 3) = 0$.$t = 0$ or $t^2 = 0.75 \Rightarrow t \approx 0.866$ s.So statement A is correct.Turning point when velocity $v = \frac{dx}{dt} = 12t^2 - 3 = 0$.$t^2 = 0.25 \Rightarrow t = 0.5$ s.Position: $x(0.5) = 4(0.5)^3 - 3(0.5) = 0.5 - 1.5 = -1$ m.Distance from origin = $|{-1}| = 1$ unit.So statement B is correct, D is incorrect.Acceleration $a = \frac{dv}{dt} = 24t$. For $t \ge 0$, $a \ge 0$.Statement C is correct.Since the particle turns at $t = 0.5$ s, statement E is incorrect.Answer:Correct statements: A, B, C. Option B matches.Option (B) A, B, C Only