A particle starts moving from time t equals zero and its coordinate is given as x of t equals 4t cube minus 3t

Q. A particle starts moving from time t = 0 and its coordinate is given as x(t) = 4t³ − 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

Correct Answer: A, B, C Only

Solution

The position of the particle is given by:

x(t) = 4t³ − 3t

Velocity is the first derivative of position:

v(t) = dx/dt = 12t² − 3

At turning point, velocity is zero:

12t² − 3 = 0 ⇒ t² = 1/4 ⇒ t = 0.5

Position at turning point:

x(0.5) = 4(0.125) − 3(0.5) = 0.5 − 1.5 = −1

Hence the particle is 1 unit away from the origin, so statement B is correct and D is incorrect.

Acceleration is the second derivative of position:

a(t) = d²x/dt² = 24t

For t ≥ 0, acceleration is non-negative. Hence statement C is correct.

To find when the particle returns to origin:

4t³ − 3t = 0 ⇒ t(4t² − 3) = 0 ⇒ t = √3/2 ≈ 0.866

Hence statement A is correct.

Therefore, the correct statements are A, B and C only.