Explain clearly, with examples, the distinction between
$(a)$ magnitude of displacement 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].

(N/A) The magnitude of displacement over an interval of time is the shortest distance (a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual distance covered by the particle in a given interval of time.
For example, consider a particle moving from point $A$ to point $B$ and then returning to point $C$, taking a total time $t$. As shown in the figure, the magnitude of displacement is $AC$, whereas the total path length is $AB + BC$.
Note that the magnitude of displacement can never be greater than the total path length. However, in cases where the particle moves in a single direction without turning back, both quantities are equal.
$(b)$ Magnitude of average velocity = $\frac{\text{Magnitude of displacement}}{\text{Time interval}}$
For the given particle, Average velocity = $\frac{AC}{t}$.
Average speed = $\frac{\text{Total path length}}{\text{Time interval}} = \frac{AB + BC}{t}$.
Since $(AB + BC) > AC$, the average speed is greater than the magnitude of average velocity. The two quantities are equal if the particle continues to move along a straight line in the same direction.