Distance is the area under the velocity-time graph

Steps:

• Identify the portion of the graph for the first 2 seconds.
• Calculate the area under the curve from t=0 to t=2 s, which represents displacement or distance if velocity is positive.
• Use geometry: if triangular, area = (1/2) × base × height; if rectangular or trapezoidal, adjust accordingly.
• The diagram's specific shape yields 28 m.

Not enough information: Without the actual diagram, precise calculation is impossible.

Why D is correct:

• Area under v-t graph equals distance travelled (definition of displacement from integration of velocity).

Why the others are wrong:

• A: Underestimates area, perhaps confusing with average velocity times time incorrectly.
• B: Far too small, likely misreading scale or units.
• C: Possible miscalculation of full graph area instead of first 2 s.

Final answer: D