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MCQMediumJEE 2025Graphs of Motion

JEE Physics 2025 Question with Solution

Which of the following curves possibly represent one-dimensional motion of a particle?

  • A
    Graph of phase versus time with straight line of positive slope passing through the origin on labeled phase and time axes.
  • B
    Graph of velocity versus displacement showing a closed circular curve centered at the origin on labeled velocity and displacement axes.
  • C
    Graph of velocity versus time showing a closed circular curve centered at the origin on labeled velocity and time axes.
  • D
    Graph of total distance versus time showing distance increasing, then remaining constant briefly, and then increasing again on labeled axes.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Four different curves are shown.

Find: Which curves can possibly represent one-dimensional motion of a particle, and hence the correct option.

To test whether a graph can represent physical motion, check whether the plotted quantity is physically consistent with one-dimensional kinematics.

  • Any quantity plotted against time must be single-valued at each instant of time.
  • Total distance can never decrease with time.
  • For SHM, phase can vary linearly with time, and the vv-xx relation can form an ellipse or a circle under suitable scaling.

For curve (A), phase versus time is a straight line. In SHM, phase is

ϕ(t)=ωt+ϕ0\phi(t)=\omega t+\phi_0

So a linear phase-time graph is physically possible. Hence (A) is possible.

For curve (B), velocity versus displacement is a closed curve. In SHM,

x2A2+v2(Aω)2=1\frac{x^2}{A^2}+\frac{v^2}{(A\omega)^2}=1

This is an ellipse, and with suitable scaling it can appear as a circle. Therefore (B) is also possible.

For curve (C), velocity versus time is a circle. A vertical line at a given time cuts the curve at two points, implying two different velocities at the same instant. That is physically impossible. Hence (C) is not possible.

For curve (D), total distance versus time increases, then remains constant, then increases again. Total distance is always non-decreasing, and a flat portion means the particle is at rest for some time. Hence (D) is possible.

Therefore, the possible curves are (A), (B), and (D) only.

Among the given combinations, this corresponds to option (1). The correct option is A.

Quick Graph Validity Check

Given: Four graphs representing different physical quantities.

Find: Which ones can represent one-dimensional motion.

Use a quick screening rule:

  1. If a graph is plotted against time, it must give only one value at each time.
  2. Distance versus time can stay constant or rise, but cannot fall.
  3. In oscillatory motion, phase versus time can be linear and vv versus xx can be a closed ellipse or circle-like curve.

Applying this:

  • (A) valid
  • (B) valid
  • (C) invalid because one time gives two velocities
  • (D) valid because total distance never decreases

So the correct set is A, B and D only, which corresponds to option A.

Common mistakes

  • Students often reject the vv versus xx closed curve in (B) because one displacement corresponds to two velocities. This is not wrong: at the same displacement, the particle may move in opposite directions at different instants. The uniqueness requirement applies to a quantity versus time, not necessarily versus another variable like displacement.

  • A common mistake is to accept (C) as valid by treating it like a phase-space curve. This is wrong because the horizontal axis is time, so each instant must correspond to exactly one velocity. Use the vertical line test for graphs against time.

  • Some students think total distance in (D) must always increase strictly. This is incorrect because the particle may remain at rest for some interval, making the graph horizontal. The correct condition is that total distance must be non-decreasing, not necessarily strictly increasing.

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