A particle starts moving from time t = 0 and its coordinate is given… | JEE Main 2026 (28 Jan, Shift 2)

Q1JEE Main 2026 · 28 JAN · SHIFT 2Medium

Velocity, Acceleration & Variable Motion

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:

Show answer and solution

Answer: (b)

A: $x = 0 \Rightarrow 4t^2 = 3 \Rightarrow t = \dfrac{\sqrt{3}}{2} = 0.866$ , true.
Turning point: $v = 12t^2 - 3 = 0 \Rightarrow t = 0.5$ ; $x = 0.5 - 1.5 = -1$ , so B true, D false.
C: $a = 24t \ge 0$ for $t \ge 0$ , true.
E: the particle does turn back at $t = 0.5$ s, false.
So the answer is (b) A, B, C only.

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Velocity, Acceleration & Variable Motion question from JEE Main 2026 (28 Jan, Shift 2)

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