Question

Describe with examples clearly, the distinction between:
(a) Magnitude of displacement otherwise known as distance over an interval of time, and the total length of path traversed by a particle over the identical interval.
(b) Magnitude of average velocity over an interval of time, and the average speed over the similar interval. [Average speed of a particle over an interval of time is called the total path length taken divided by the time interval].
Represent in both (a) and (b) that the second quantity is either greater than or equal to the first. When will this equality sign be true? [Let us consider one-dimensional motion only].

Answer 1

Displacement is a vector quantity depending on both the magnitude and direction. It is to be noted that the magnitude of displacement will not never be greater than the total path length. Average speed of a quantity is defined as the ratio of the total distance of the body to the total time taken. This will help you in answering this question.

Complete step by step answer:
(a) The shortest distance, more likely to be a straight line between the initial and final positions of the particle will be the magnitude of displacement over an interval of time. The total path length of a particle will be the original path length traversed by the particle in a certain interval of time. Let us take an example. Consider a particle moving from point $A$ to point $B$ and then, bounce back to a point, $C$ taking a total time, given in the diagram.
Then, the magnitude of displacement of the particle will be equivalent to $AC$.
Hence,
$\text{total path length=AB+BC}$
It is also necessary to note that the magnitude of displacement can never be greater than the total path length. Both quantities can be equal to each other in some situations.
(b) Magnitude of average velocity will be,
$\left| {{v}_{avg}} \right|=\dfrac{\left| d \right|}{t}$
Where $\left| d \right|$ be the magnitude of displacement and $t$ be the time interval
For a specific particle,
Average velocity is,
${{v}_{avg}}=\dfrac{AC}{t}$
Average speed is the ratio of total path length to the time interval
${{s}_{avg}}=\dfrac{AB+BC}{t}$
As we can see, $AB+BC > AC$ average speed is higher than the magnitude of average velocity. The two quantities will be equivalent when the particle continues to move along a straight line. Therefore the question has been answered.

Note: The velocity is the time rate of variation of the displacement. Distance covered is the total length of the path traversed by a body. The displacement is the shortest distance between the initial and final points. Time rate of variation of distance is known as speed.