JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

A particle starts moving from time $t = 0$ and its coordinate is given as $x (t) = 4 t^{3} - 3 t$ .

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

Solution

$x (t) = 4 t^{3} - 3 t$ , $v (t) = 12 t^{2} - 3$ , $a (t) = 24 t$ .

A. $x = 0 \Rightarrow t (4 t^{2} - 3) = 0 \Rightarrow t = 0$ or $t = \frac{3}{2} \approx 0.866$ . So returns to origin at $t \approx 0.866$ — TRUE.

B / D. Turning point: $v = 0 \Rightarrow t = \frac{1}{2}$ . Then $x = 4 \cdot \frac{1}{8} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = - 1$ . So $∣ x ∣ = 1$ — B is TRUE, D is FALSE.

C. $a = 24 t \geq 0$ for $t \geq 0$ — TRUE.

E. Initial velocity $v (0) = - 3 < 0$ , so the particle moves in $- x$ direction initially, then turns back at $t = 1/2$ — the particle does turn back. FALSE.

So A, B, C only are correct.

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