JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Kinematics

Reading displacements from the graph, the line from $t=1$ s to $t=3$ s rises straight from $-5$ m to $5$ m, so its slope is $\frac{5-(-5)}{3-1}=5$ m/s. The average velocity from $0$ to $3$ s is $\frac{5-0}{3}=\frac{5}{3}$ m/s, not $10$ m/s, so that claim fails. Between $3$ s and $5$ s the displacement stays at $5$ m, giving an average velocity $\frac{5-5}{2}=0$, which is correct. At $t=2$ s the instantaneous velocity is the slope of the line through that instant, $5$ m/s, which is correct. From $5$ s to $7$ s the displacement falls from $5$ m to $0$, an average velocity of $-2.5$ m/s, while at $t=6.5$ s the slope of the steep fall from $10$ m to $0$ is $-10$ m/s, so these two are unequal and that claim fails. Over $0$ to $9$ s the displacement returns to zero, making the average velocity $\frac{0-0}{9}=0$, which is correct. The correct statements are (B), (C) and (E).