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Explain clearly, with examples, the distinction between:(a) 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;(b) 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]. |
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Answer» (a) Consider a person who leaves home for work in the morning and returns home in the evening. The magnitude of the displacement of the person during the interval is zero but the total length of path covered = 2 × distance between the home and the workplace. Magnitude of displacement is the magnitude of the shortest length between the two points. The second quantity is clearly always greater than or equal to the first quantity. The equality holds if the body has not moved at all (path length = magnitude of displacement = 0) (b) Average velocity = Δx/Δt, is zero (like in the example explained in) (a) then average velocity = 0. Average speed = Total path length/Time taken 2 × (Distance between the home and the work place/Time taken to traverse this distance) = Clearly average speed ≥ magnitude of average velocity. Equality exists when body does not undergo any motion. |
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