JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Kinematics

The total distance is the sum of the magnitudes of the areas between the velocity graph and the time axis. From $t=0$ to $t=20\text{s}$ the speed rises to $5\text{m/s}$ and returns to zero, enclosing a triangle of area $\frac{1}{2}(20)(5)=50\text{m}$; from $t=20\text{s}$ to $t=40\text{s}$ the velocity is negative with an equal triangular area $\frac{1}{2}(20)(5)=50\text{m}$. The total distance travelled is $50+50=100\text{m}$. The displacement is the signed sum of the areas, $+50-50=0$, so the particle returns to its starting point, and the average velocity is $\frac{\text{displacement}}{\text{time}}=\frac{0}{40}=0$. The total distance is $100\text{m}$ and the average velocity is zero.