The velocity ( v ) versus time ( t ) plot of a particle is shown in...

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:
$A_1 = \dfrac{1}{2} \times 20 \times 5 = 50\text{ m}$

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

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 $= \dfrac{\text{Total displacement}}{\text{Total time}} = \dfrac{0}{40} = 0$

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