Question

A particle starts moving from time \(t=0\) and its coordinate is given as \[ x(t)=4t^3-3t. \] Consider the following statements:
[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.] The particle never turns back as acceleration is non-negative.
Choose the correct answer from the options given below:

• A. C, E Only
• B. A, B, C Only
• C. A, C, D Only
• D. A, C Only

Hint:

Turning points depend on velocity becoming zero, not on acceleration alone.

Answer 1

The Correct Option is C

Step 1: Velocity and acceleration \[ v(t)=\frac{dx}{dt}=12t^2-3 \] \[ a(t)=\frac{dv}{dt}=24t \] Since \(t\ge 0\), \[ a(t)\ge 0 \] Hence, statement C is true.
Step 2: Turning point Turning point occurs when velocity is zero: \[ 12t^2-3=0 \Rightarrow t=\frac12 \] Position at turning point: \[ x\left(\frac12\right) =4\left(\frac18\right)-3\left(\frac12\right) =\frac12-\frac32=-1 \] Distance from origin: \[ |x|=1 \] Hence:
Statement B is true.
Statement D is false.

Step 3: Return to origin Set \(x(t)=0\): \[ 4t^3-3t=0 \Rightarrow t(4t^2-3)=0 \] Non-zero solution: \[ t=\sqrt{\frac34}=\frac{\sqrt3}{2}\approx0.866 \] Thus, statement A is true.
Step 4: Check statement E Even though acceleration is non-negative, velocity changes sign at \(t=\frac12\), so the particle does turn back. Hence, statement E is false. Conclusion: Correct statements are: \[ \boxed{A,\ C,\ D} \] Final Answer: \[ \boxed{\text{(C) A, C, D Only}} \]