JEE Physics 2026 Question with Solution

Given: $x(t)=4t^3-3t$ for $t\ge 0$.

Find: Which statements among $A, B, C, D, E$ are correct.

Step 1: Velocity and acceleration

$$v(t)=\frac{dx}{dt}=12t^2-3$$ $$a(t)=\frac{dv}{dt}=24t$$

Since $t\ge 0$,

$$a(t)\ge 0$$

Hence, statement C is true.

Step 2: Turning point Turning point occurs when velocity is zero:

$$12t^2-3=0$$ $$t=\frac12$$

Position at turning point:

$$x\left(\frac12\right)=4\left(\frac18\right)-3\left(\frac12\right)=\frac12-\frac32=-1$$

Distance from origin:

$$|x|=1$$

Hence, statement B is true and statement D is false.

Step 3: Return to origin Set $x(t)=0$:

$$4t^3-3t=0$$ $$t(4t^2-3)=0$$

Non-zero solution:

$$t=\sqrt{\frac34}=\frac{\sqrt3}{2}\approx 0.866$$

Thus, statement A is true.

Step 4: Check statement E Even though acceleration is non-negative, velocity changes sign at $t=\frac12$, so the particle does turn back. Hence, statement E is false.

Conclusion: The working shows the correct statements are A, B, C. Therefore, the defensible answer from the given options is B. The solution incorrectly lists A, C, D and option C, which disagrees with its own calculation that the turning-point distance is $1$, not $0.5$.

Given: $x(t)=4t^3-3t$.

Find: The correct option by checking each statement directly.

From

$$v(t)=12t^2-3$$

we see that for $0\le t<\frac12$,

$$v(t)<0$$

and for $t>\frac12$,

$$v(t)>0$$

So the particle first moves in the negative direction, comes to rest at $t=\frac12$, and then reverses direction. Therefore, it has a turning point.

At the turning point,

$$x\left(\frac12\right)=-1$$

so the distance from origin is $1$. Hence B is true and D is false.

Also,

$$a(t)=24t\ge 0 \quad \text{for} \quad t\ge 0$$

so C is true.

For return to origin,

$$4t^3-3t=0 \Rightarrow t=0 \text{ or } t=\frac{\sqrt3}{2}$$

Thus after starting, it returns to origin at

$$\frac{\sqrt3}{2}\approx 0.866$$

so A is true.

Therefore, the true set is A, B, C, not A, C, D. The correct option should be B based on the actual working.