### Related Formula For realistic physical motion in one dimension: 1. Time t$t$ can never be negative during a normal positive time sequence, and cannot flow backwards. 2. Total distance covered can never decrease over time. 3. A particle cannot have two different values of position or velocity at the exact same instant of time. ### Core Logic Let us analyze each curve: - **Curve (A)** (Phase phi$\phi$ vs Time t$t$): Represents phi = kt + C$\phi = kt + C$, which is a valid linear relationship of phase over time (e.g., in Simple Harmonic Motion x = Asin(kt + C)$x = A\sin(kt + C)$). (Valid) - **Curve (B)** (Velocity v$v$ vs Displacement x$x$): A closed loop, which represents symmetric harmonic-type oscillation. For example, v^2 + omega^2 x^2 = textconst$v^2 + \omega^2 x^2 = \text{const}$ (ellipse) is a perfectly physically valid 1D SHM velocity-displacement phase portrait. (Valid) - **Curve (C)** (Velocity vs Time): The curve enters into the negative time quadrant. Time cannot go backwards or exist in negative values relative to starting sequence in standard physical scenarios. (Invalid) - **Curve (D)** (Total Distance vs Time): Represents total distance increasing over time. Total distance is a non-decreasing function of time (d(d)/dt ge 0$d(d)/dt \ge 0$). Thus, this curve is physically valid. (Valid) ### Step 1: Conclusion Therefore, curves A, B, and D possibly represent physical one-dimensional motion. The correct option is (1). ### Pattern Recognition Quick check for graph validity: 1. Time cannot run backwards (ruling out C). 2. Total distance can never decrease (D is valid because it strictly goes upwards). 3. v$v$ vs x$x$ can be circular/elliptical in SHM (B is valid). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line