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].

During the first $18\,min$ of a $60\,min$ trip, a car has an average speed of $11\,m / s$. What should be the average speed for remaining $42\,min$ so that car is having an average speed of $21\,m / s$ for the entire trip?

A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the partk. Then cycles along the circumference and returns to the centre along $QO$ as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour respectively) is

A car goes from $A$ to $B$ (without changing direction), during first part of journey its average speed is $V_1$ and for second part of journey average speed is $V_2$ . The ratio path length of second part to the path length of first part is $\sqrt {\frac{{{V_2}}}{{{V_1}}}} $ . In that case the average speed of total path is ……. of average speeds of both parts. Choose correct option of fill. 

Which among average speed and average velocity can be positive, negative or zero ?