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

$A$ person walks along a straight road from his house to a market $2.5 \,km$ away with a speed of $5 \,km/h$ and instantly turns back and reaches his house with a speed of $7.5 \,km/h$. The average speed of the person during the time interval $0$ to $50 \,min$ is (in $m/s$)

An object moves with speeds $v_1, v_2$,and $v_3$ along line segments $AB, BC$,and $CD$ respectively,as shown in the figure. If $AB = BC$ and $AD = 3 AB$,then the average speed of the object will be:

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:

$A$ 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 = $ . . . . . . $m/s$.

$A$ car starts from rest and moves such that it possesses average speeds of $5 \, ms^{-1}$,$10 \, ms^{-1}$,and $15 \, ms^{-1}$ in the first,second,and third seconds,respectively. What is the total distance covered by the car in these three seconds? ......... $m$