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Question
Explain clearly, with examples, the distinction between:
- magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.
- magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Solution
- Displacement magnitude of a particle in motion for a certain period is the shortest distance between its start and end points during that time, while the total path length is the actual distance traveled by the particle. If a particle moves from A to B and then B to C as shown,
The displacement magnitude is equal to the distance AC. The total path length = distance AB + distance BC. From the diagram, we see that the total path length (AB + BC) is greater than the displacement magnitude (AC). If the motion of the particle is in one dimension, i.e., along a straight line, then the total path length equals the displacement magnitude for the given time. - Magnitude of average velocity = `"Magnitude of displacement"/"Time interval"`
For the given particle.
Average velocity = `("AC")/"t"`
Average speed = `"Total path length"/ "Time interval"`
`=("AB"+"BC")/"t"`
Since the total path length (AB + BC) exceeds the displacement (AC), the average speed of the particle is greater than its average velocity. If the particle travels in a straight line, the displacement over a certain period is the same as the path it actually covers. Thus, in this scenario, the average speed is equivalent to the average velocity.
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