$A$ 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 magnitude of average velocity in $m/s$?

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].

$A$ particle moves in the east direction with $15 \, m/s$ for $2 \, s$,then moves northward with $5 \, m/s$ for $8 \, s$. The average velocity of the particle is:

$A$ car moves from $X$ to $Y$ with a uniform speed $v_1$ and returns from $Y$ to $X$ with a uniform speed $v_2$. The average speed for this round trip is

If a car travels $40 \%$ of the total distance with a speed $v_1$ and the remaining distance with a speed $v_2$,then the average speed of the car is:

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