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 :

Solution

Physics sir · 4 steps

Differentiate twice: v=0 gives the turning point, and the sign of a tells you how it got there.

1. Velocity and acceleration. From x = 4t³ - 3t, the velocity is v = 12t² - 3 and the acceleration is a = 24t. At t = 0 the velocity is -3, so the particle starts by moving towards negative x.
2. Turning point. The particle turns back when v = 0, that is 12t² = 3, so t = 0.5. There x = 4(0.125) - 1.5 = -1, which is 1 unit from the origin.
3. Back to origin. Set x = 0 for t > 0: 4t² = 3, so t = √3/2 ≈ 0.866.

   Statement	Check	Verdict
   A	x = 0 again at t = 0.866	true
   B	| x | = 1 at the turning point t = 0.5	true
   C	a = 24t ≥ 0 for all t ≥ 0	true
   D	0.5 is the time of turning, not the distance	false
   E	v changes sign at t = 0.5, so it does turn back	false

4. So. A, B and C are correct, which is option (b).

The idea to keep. A particle turns back where velocity changes sign, not where acceleration does. Positive acceleration on a particle moving backwards first slows it, stops it, then reverses it.

Answer from the NTA final answer key. Solution by PrepWiser.