The figure is a velocity–time graph. The quantity you probably want is the displacement (net) from t = 0 to t = t2, which is the area under the curve between those times.

Key idea:
- Net displacement s(0→t2) = ∫ from 0 to t2 of V dt = area of the positive lobe (0 to t1) plus the area of the negative lobe (t1 to t2).
- If you want the distance travelled (not signed), take the sum of the absolute areas: D = A_positive + |A_negative|.

If you approximate the lobes as triangles (as the grid suggests):
- Positive lobe (0 to t1): A_pos ≈ (1/2) * Vmax * t1, where Vmax is the maximum velocity.
- Negative lobe (t1 to t2): A_neg ≈ -(1/2) * |Vmin| * (t2 − t1), where |Vmin| is the magnitude of the minimum velocity.

Then:
- Net displacement: s(0→t2) ≈ (1/2) Vmax t1 − (1/2) |Vmin| (t2 − t1).
- Distance travelled up to t2: D ≈ (1/2) Vmax t1 + (1/2) |Vmin| (t2 − t1).

If you provide the actual numeric values (Vmax, |Vmin|, t1, t2) or the velocity values at the grid points, I can compute the exact numbers.