JEE Main 2026 — Motion In One Dimension Question with Solution

The total distance travelled by the particle is equal to the sum of the magnitudes of the areas under the velocity-time graph.

Area of the first triangle (from $t=0$ to $t=20\text{ s}$) is:
$A_1=\frac{1}{2}\times 20\times 5=50\text{ m}$

Magnitude of the area of the second triangle (from $t=20$ to $t=40\text{ s}$) is:
$|A_2|=\left|\frac{1}{2}\times 20\times (-5)\right|=50\text{ m}$

Total distance $=A_1+|A_2|=50+50=100\text{ m}$

The total displacement is the algebraic sum of the areas under the velocity-time graph.

Total displacement $=A_1+A_2=50-50=0\text{ m}$

Average velocity is the ratio of total displacement to total time.

Average velocity $=\frac{\text{Total displacement}}{\text{Total time}}=\frac{0}{40}=0$

Answer: $100\text{ m}$ and zero