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

To determine the correct statements about the motion of the particle, we will analyze each statement based on the given coordinate function \( x(t) = 4t^3 - 3t \).

1. Statement A: The particle returns to its original position (origin) 0.866 units later.
   • For the particle to return to the origin, we set \( x(t) = 0 \). Therefore: \(4t^3 - 3t = 0\)
   • Factoring gives: \(t(4t^2 - 3) = 0\)
   • This implies: \(t = 0 \text{ or } 4t^2 - 3 = 0\)
   • Solving \( 4t^2 = 3 \) gives: \(t^2 = \frac{3}{4} \implies t = \pm\frac{\sqrt{3}}{2} \approx \pm0.866\)
   • Thus, the particle returns to the origin at \( t \approx 0.866 \) (since negative time is not physical, we consider the positive root).
2. Statement B: The particle is 1 unit away from the origin at its turning point.
   • To find the turning point, differentiate \( x(t) \) to get velocity: \(v(t) = \frac{dx}{dt} = 12t^2 - 3\)
   • Setting velocity to zero to find turning points: \(12t^2 - 3 = 0 \implies 12t^2 = 3 \implies t^2 = \frac{1}{4}\)
   • Solves to: \(t = \pm 0.5\) (here we consider positive \( t \) only)
   • Substitute this \( t \) back in \( x(t) \): \(x(0.5) = 4(0.5)^3 - 3(0.5) = 1 - 1.5 = -0.5\)
   • The particle is 0.5 units away, not 1 unit.
3. Statement C: Acceleration of the particle is non-negative.
   • Acceleration is the derivative of velocity: \(a(t) = \frac{d^2x}{dt^2} = 24t\)
   • Since time \( t \geq 0 \), acceleration \( a(t) \geq 0 \).
4. Statement D: The particle is 0.5 units away from the origin at its turning point.
   • As calculated in Statement B, at the turning point \( t = 0.5 \): \(x(0.5) = -0.5\)
   • Thus, the particle is indeed 0.5 units from the origin.
5. Statement E: The particle never turns back as acceleration is non-negative.
   • This is incorrect because a turning point (direction change) is determined by velocity becoming zero, not by non-negative acceleration.
   • As shown, there is a turning point at \( t = 0.5 \).

Based on the analysis, the correct answer is the combination of statements A, C, and D.