A particle moving in a straight line covers half the distance with speed $6 \mathrm{~m} / \mathrm{s}$. The other half is covered in two equal time intervals with speeds 9 $\mathrm{m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$ respectively. The average speed of the particle during the motion is :

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].

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 (in $km/h$) of the man over the interval of time $0$ to $50\; min$ ?

A body is moving with variable acceleration $(a)$ along a straight line. The average acceleration of body in time interval $t_1$ to $t_2$ is

''The magnitude of average velocity is equal to average speed''. This statement is not always correct and not always incorrect. Explain with example.

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