Explain clearly, with examples, the distinction between 

$(a)$ magnitude of displacement (somettmes 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].

The magnitude of displace 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. Ctaking a total time (q. as shown below. Then, the megnitude of displacement

of the particle $=A C$

Whereas, total path length = $A B+B C$

pnote 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.

(b) Magnitude of average velocity = Magnitude of displacement/Time interval

For the given particle.

Average velocity $=A C / t$ Average speed = Total path length/Time interval $=(A B+BC) / t$

since, $(A B+B C) > A C$ average speedis greater than the magnitude of average velocity. The two quantities will be equalif the particle continues to move along a straight