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 :

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) A, B, C only

Step-by-step Solution

Detailed explanation

\(x=0\Rightarrow t=0,\frac{\sqrt{3}}{2}\)
\(v=12t^{2}-3\)
At turning point, \(v=0\)
\(t=\frac{1}{2}\Rightarrow x=\frac{4}{8}-\frac{3}{2}=-1\)
\(a=24t\) (always positive)

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