JEE PYQ: Motion in a Straight Line - Question ID 2b8b065cdd64 (JEE Main 2026)

Core Formula & Concept:

In kinematics of one-dimensional motion, the position of a particle is given by a function of time $x(t)$. From this we derive:

• Velocity $v(t) = \frac{dx}{dt}$
• Acceleration $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
• A turning point occurs where the velocity changes sign, i.e., $v(t) = 0$.
• The particle “returns to the origin” when $x(t) = 0$ at some $t > 0$.

Step-by-Step Derivation:

Step 1: Find when the particle returns to the origin.

Set $x(t) = 0$: $4t^3 - 3t = 0 \quad\Longrightarrow\quad t(4t^2 - 3) = 0.$ Nonzero roots are $t = \pm\sqrt{\tfrac{3}{4}} = \pm\frac{\sqrt{3}}{2}$. The positive root is $t = \frac{\sqrt{3}}{2} \approx 0.866.$ Hence statement A (“0.866 units later”) is correct.

Step 2: Locate the turning points.

Velocity $v(t) = \frac{dx}{dt} = 12t^2 - 3$. Set $v(t) = 0$: $12t^2 - 3 = 0 \quad\Longrightarrow\quad t^2 = \tfrac{1}{4} \quad\Longrightarrow\quad t = \pm\tfrac{1}{2}.$ The positive turning point is at $t = \tfrac{1}{2}$. Compute $x(\tfrac{1}{2})$: $x(\tfrac{1}{2}) = 4\bigl(\tfrac{1}{2}\bigr)^3 - 3\bigl(\tfrac{1}{2}\bigr) = 4\cdot\tfrac{1}{8} - \tfrac{3}{2} = \tfrac{1}{2} - \tfrac{3}{2} = -1.$ The distance from the origin is $|x| = 1$. Thus statement B (“1 unit away”) is correct, while statement D (“0.5 units away”) is incorrect.

Step 3: Examine the acceleration.

Acceleration $a(t) = \frac{dv}{dt} = 24t$. For $t \ge 0$, $a(t) \ge 0$. Therefore statement C (“acceleration is non-negative”) is correct.

Step 4: Check whether the particle ever turns back.

The velocity $v(t) = 12t^2 - 3$ changes sign at $t = \tfrac{1}{2}$, so the particle does reverse direction. Statement E (“never turns back”) is incorrect.

Conclusion:

Correct statements are A, B, and C. The matching option is B.

Common Traps & Exam Tip:

1. Sign errors in roots: Students often forget the negative root $t = -\tfrac{\sqrt{3}}{2}$, but only the positive root matters for “returns later.” 2. Confusing position with distance: At the turning point, $x = -1$, so the distance is $1$, not $0.5$. 3. Acceleration misinterpretation: Non-negative acceleration does not prevent velocity from changing sign; it only means the acceleration never points opposite to the positive $x$-axis.