Explain clearly, with examples, the distinction between 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 1
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 (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.
Solution 2
Suppose, in the above example, the particle takes time t to cover the whole journey. Then, the magnitude of the average velocity of the particle over time-interval t is = Magnitude of displacement /Time-interval =0/t =0
While the average speed of the particle over the same time-interval is =Total path length /Time-interval= 2 AB /t
Again, the second quantity (average speed) is greater than the first (magnitude of average velocity).
