$A$ man walks on a straight road from his home to a market $2.5 \; km$ away with a speed of $5 \; km \; h^{-1}$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \; km \; h^{-1}$. What is the average speed of the man over the interval of time $0$ to $30 \; min$?

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].

$A$ body moves with velocities $v$,$2v$,and $3v$ in the first,second,and third one-third distances of the path traveled,respectively. Its average speed is:

Is the statement "$A$ particle can have zero speed but non-zero velocity" true or false? Explain.

The ratio of the numerical values of the average velocity and average speed of a body is always

$A$ particle covers half of its total distance with speed $v_1$ and the rest half distance with speed $v_2$. Its average speed during the complete journey is