Velocity-time Graph Interpretation

Question

For a particle with a velocity-time graph forming a straight line from 10 m/s at t=0 to 22 m/s at t=6 s, what is the displacement during this interval?

RankGuru · Accepted Answer

1. **Graphical Principle**: Displacement is represented by the area under the velocity-time graph.
2. **Feature Analysis**: Since the graph is a straight line, the acceleration is uniform, and the area forms a trapezoid.
3. **Method 1: Average Velocity**:
   - $v_{avg} = \frac{u + v}{2} = \frac{10 + 22}{2} = 16$ m/s.
   - Displacement $s = v_{avg} 	imes t = 16 	imes 6 = 96$ m.
4. **Method 2: Area Calculation**:
   - Area of trapezoid $= \frac{1}{2} (h_1 + h_2) 	imes w = \frac{1}{2} (10 + 22) 	imes 6 = 96$ m.
5. **Conclusion**: The total displacement is **$96$ m**.

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Velocity-time Graph Interpretation

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