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

Question

A particle starts moving from time $t = 0$ and its coordinate is given as $x (t) = 4 t^{3} - 3 t$ . [2026]

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:

Smart Achievers · Accepted Answer

(2)

$x (t) = 4 t^{3} - 3 t$

For $x (t) = 0 \Rightarrow t (4 t^{2} - 3) = 0$

$\Rightarrow t = \frac{\sqrt{3}}{2} \approx 0.866 units$

$v (t) = 12 t^{2} - 3$

Turning point at $v = 0$

$t^{2} = \frac{1}{4} \Rightarrow t = 0.5 units$

At $t = \frac{1}{2}$ unit, $x (\frac{1}{2}) = 4 {(\frac{1}{2})}^{3} - 3 (\frac{1}{2}) = - 1.0 unit$

$a (t) = 24 t$

which is non-negative for $t \geq 0$ .

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