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 onedimensional motion only).

Explain clearly, with examples, the distinction between : (a) magnitud

1.	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 onedimensional motion only).
Answer» Solution :(a) Displacement is SHORTEST distance between FINAL POSITION and initial position for moving object. Path length is total distance travelled b y object. Suppose, object goes to B from A in .t. time, the displacement = AB and path length = AB `therefore` Displacement = Path length Now if object goes grom A to B and then to C, then displacement AC = AB-BC and path length = AB + BC. `therefore` Displacement `lt ` Path length and average velocity `(AC)/(t)lt ` average speed `(AB+BC)/(t)` Thus, average velocity `lt` average speed.