JEE Mains · Physics · STD 11 - 2. motion in straight line
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 :
- A A, C, D only
- B A, B, C only
- C C, E only
- D A, C only
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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