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

Question

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

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 : [2026]

Smart Achievers · Accepted Answer

(3)

$x = 0 \Rightarrow t = 0, \frac{\sqrt{3}}{2}$

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

$At turning point, v = 0$

$t = \frac{1}{2} \Rightarrow x = \frac{4}{8} - \frac{3}{2} = - 1$

$a = 24 t (always positive)$

Solution

1 A, C, D Only

2 A, C Only

3 A, B, C Only

4 C, E Only

Ans.
(3)

$x = 0 \Rightarrow t = 0, \frac{\sqrt{3}}{2}$

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

$At turning point, v = 0$

$t = \frac{1}{2} \Rightarrow x = \frac{4}{8} - \frac{3}{2} = - 1$

$a = 24 t (always positive)$

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