Given x(t)=4t3−3t, we find v(t)=12t2−3 and a(t)=24t.
Statement A: Setting x(t)=0 gives 4t3−3t=0⇒t(4t2−3)=0, so t=0 or t=3/4=0.866. The particle returns to origin at t=0.866 units. TRUE.
Statement B: Turning points occur at v(t)=0⇒12t2−3=0⇒t=±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≥0 (particle starts at t=0), a(t)=24t≥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≥0, and particle does turn back at t=0.5. FALSE.
Correct statements: A, B, C Only.