Question

Explain clearly, with examples, the distinction between : 

1. 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; 
2. 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].

Answer 1

a. The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC. 

Whereas, total path length = AB + BC 
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.

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b. Magnitude of average velocity = \(\frac{\text{Magnitude of displacement }}{\text {Time interval}}\)
For the given particle , 
Average velocity = \(\frac{\text {AC}}{\text t}\)
Average speed = \(\frac{\text{Total path length }}{\text{ Time interval}}\) 
=\(\frac{\text {AB}+\text{BC}}{\text t}\)
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.