Velocity-time Graph Interpretation
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?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- velocity-time graph interpretation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
96 m
- Graphical Principle: Displacement is represented by the area under the velocity-time graph.
- Feature Analysis: Since the graph is a straight line, the acceleration is uniform, and the area forms a trapezoid.
- Method 1: Average Velocity:
- vavg=2u+v=210+22=16 m/s.
- Displacement s=vavg×t=16×6=96 m.
- Method 2: Area Calculation:
- Area of trapezoid =21(h1+h2)×w=21(10+22)×6=96 m.
- Conclusion: The total displacement is 96 m.
This medium difficulty physics question is from the chapter kinematics, covering the topic of velocity-time graph interpretation. It appeared in the 2025 exam.
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