Which of the following curves possibly represent one-dimensional motion of a particle?
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

Ask one thing of each graph: could a real particle moving on a line actually produce it?

1. The test. A real motion has exactly one velocity at each instant, time only moves forward, and total distance can never decrease. Check every graph against these rules.

   Graph	What it shows	Possible?
   (A) phase vs time	Straight line, φ = ω t + φ₀ with a negative φ₀, as in SHM	Yes
   (B) velocity vs displacement	SHM gives v² = ω²(A² - x²), an ellipse, which is a circle with suitable scales. Clockwise is right: when v > 0, x increases	Yes
   (C) velocity vs time	A vertical line through the circle cuts it twice, so two velocities at one instant. The lower half also runs backward in time	No
   (D) total distance vs time	Never decreases. The slope is the speed, which only changes from one segment to the next	Yes

2. Why C fails. Velocity must be a single-valued function of time. A closed loop on a v–t graph breaks this, so no particle can follow it.
3. So. A, B and D are possible and C is not. The answer is (c).

The idea to keep. Any graph with time on the x-axis must give one value per instant and must never loop back. A distance–time graph can never slope downward.

Answer from the NTA final answer key. Solution by PrepWiser.