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Physics · Motion in one Dimension

JEE Main 2026 — 28 January, Evening Shift — Question 33

A particle starts moving from time t=0\mathrm{t}=0 and its coordinate is given as x(t)=4t3−3t\mathrm{x}(\mathrm{t})=4 \mathrm{t}^{3}-3 \mathrm{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:

  1. Option A:

    A,C,D only

  2. Option B:

    A,B,C only

  3. Option C:

    C,E only

  4. Option D:

    A,C only

    Correct

Answer: D

Step-by-step solution

x=0⇒t=0,32x=0 \Rightarrow t=0, \frac{\sqrt{3}}{2} v=12t2−3\mathrm{v}=12 \mathrm{t}^{2}-3 At turning point, v=0\mathrm{v}=0 t=12⇒x=48−32=−1\mathrm{t}=\frac{1}{2} \Rightarrow \mathrm{x}=\frac{4}{8}-\frac{3}{2}=-1 a=24t\mathrm{a}=24 \mathrm{t} (always positive)

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Motion in one Dimension
Topic
Graphical Interpretation of 1-D motion
A particle starts moving from time t =0 and its coordinate is given… | JEE Main 2026 PYQ with Solution · DhiX AI