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 origin at its turning point
C. Acceleration of the particle is non-negative
D. The particle is 0.5 units away from origin at its turning point
E. Particle never turns back as acceleration is non-negative
Choose the **correct** answer from the options given below :
Answer: (B) A, B, C Only
$v = \dfrac{dx}{dt} = 12t^2 - 3$, $\quad a = \dfrac{dv}{dt} = 24t$.
A: $x = 0 \Rightarrow t(4t^2-3) = 0 \Rightarrow t = \dfrac{\sqrt3}{2} = 0.866$ ✓
Turning point: $v = 0 \Rightarrow t = 0.5$; $x(0.5) = 4(0.125) - 1.5 = -1$, i.e. 1 unit from the origin. B ✓, D ✗
C: $a = 24t \ge 0$ for $t\ge0$ ✓
E: The particle does turn back at $t = 0.5$ (it starts with negative velocity) ✗
So A, B, C only.
Solution by Sreeraj P, M.Sc Physics