question_answer 13)

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]. (c) 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].  

Answer:

(a) Suppose a body moves from point A and to point B along a straight path and then returns back to the point A along the same path. As the body returns back to its initial position A, so magnitude of displacement = 0. Distance covered = Total length of the path covered = AB + BA = AB + AB = 2AB (b) In the above example, suppose the body takes time t to complete the whole journey. Then Magnitude of average velocity Average speed . (c) In example (a), distance covered > magnitude of displacement. In example (b), average speed > magnitude of average velocity. The sign of equality will hold when the body moves along a straight line path in a fixed direction.