The total displacement of the particle is given by the algebraic sum of the areas under the velocity-time graph. Areas above the time axis are considered positive, and areas below are considered negative.

1. Area 1 (from $t=0$ s to $t=2$ s): This segment forms a triangle above the time axis. Its base is $2$ s and its height is $4$ m/s. $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (4 \text{ m/s}) = 4 \text{ m}$

2. Area 2 (from $t=2$ s to $t=4$ s): This segment also forms a triangle above the time axis. Its base is $(4-2)=2$ s and its height is $4$ m/s. $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (4 \text{ m/s}) = 4 \text{ m}$

3. Area 3 (from $t=4$ s to $t=6$ s): This segment forms a triangle below the time axis. Its base is $(6-4)=2$ s and its height is $-2$ m/s. $A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (-2 \text{ m/s}) = -2 \text{ m}$

Total displacement $S_{\text{total}}$ is the sum of these areas: $S_{\text{total}} = A_1 + A_2 + A_3 = 4 \text{ m} + 4 \text{ m} + (-2 \text{ m}) = 6 \text{ m}$

Why other options are incorrect:

• Option A ($4$ m): This represents only the displacement during the first $2$ seconds.
• Option C ($8$ m): This would be the sum of the magnitudes of the positive displacements only ($4 ext{ m} + 4 ext{ m}$). It ignores the negative displacement.
• Option D ($10$ m): This is the total distance traveled, calculated as the sum of the magnitudes of all areas: $|A_1| + |A_2| + |A_3| = 4 ext{ m} + 4 ext{ m} + |-2 ext{ m}| = 10 ext{ m}$. Displacement considers direction, while distance does not.