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].

(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.