Displacement from area under velocity-time graph

Steps:

• Locate the last 2 seconds on the graph's time axis.
• Identify velocities: 6 m/s at the start of the interval, increasing linearly to 12 m/s at the end.
• Compute average velocity: (6 + 12)/2 = 9 m/s.
• Calculate displacement: 9 m/s × 2 s = 18 m.

Why C is correct:

• Displacement is the area under the v-t graph, here a trapezoid of 18 m² representing meters.

Why the others are wrong:

• A: Underestimates area, perhaps assuming constant 3 m/s velocity.
• B: Matches rectangle of 6 m/s × 2 s, ignoring the increase.
• D: Overestimates, possibly total graph area instead of last 2 s.

Final answer: C