$A$ person travelling on a straight line moves with a uniform velocity $v_1$ for a distance $x$ and with a uniform velocity $v_2$ for the next $\frac{3}{2} x$ distance. The average velocity in this motion is $\frac{50}{7} \ m/s$. If $v_1$ is $5 \ m/s$,then $v_2 = $ . . . . . . $m/s$.

$A$ body travelling along a straight line path travels the first half of the distance with a velocity of $7 \,m/s$. During the travel time of the second half of the distance, the first half of the time is travelled with a velocity of $14 \,m/s$ and the second half of the time is travelled with a velocity of $21 \,m/s$. Then the average velocity of the body during the entire journey is: (in $\,m/s$)

$A$ person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$. Then the relation between $v, v_1$ and $v_2$ will be:

Write the relation between speed and the magnitude of velocity.

$A$ particle moves along a straight line along the $x$-axis. Its position $(x)$ versus time $(t)$ graph is shown in the figure [$x$ in meters and $t$ in seconds]. Its average speed during this motion is (in $\text{ m/s}$)

$A$ 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 $50 \; min$?