Average Velocity and Average Speed Questions in English

(B) Speed is defined as the total distance traveled by an object per unit time interval.
Distance is a scalar quantity that represents the total path length covered by an object,and it is always non-negative (i.e.,distance $\ge 0$).
Since time is also a positive scalar quantity,the ratio of distance to time (speed) must always be non-negative.
Therefore,the speed of a moving object can never be negative.

(N/A) Uniform velocity: $A$ body is said to have uniform velocity if it covers equal displacements in equal intervals of time,regardless of how small these time intervals may be.
Non-uniform (variable) velocity: $A$ body is said to have non-uniform velocity if it covers unequal displacements in equal intervals of time,or if its direction of motion changes.

(A) Let the distance between point $A$ and point $B$ be $d$.
Time taken to travel from $A$ to $B$ is $t_1 = \frac{d}{v_1} = \frac{d}{40}$.
Time taken to travel from $B$ to $A$ is $t_2 = \frac{d}{v_2} = \frac{d}{60}$.
Total distance covered = $d + d = 2d$.
Total time taken = $t_1 + t_2 = \frac{d}{40} + \frac{d}{60} = \frac{3d + 2d}{120} = \frac{5d}{120} = \frac{d}{24}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/24} = 2 \times 24 = 48 \ km/h$.

(D) Let the total distance $AB$ be $d$.
Each segment covers a distance of $\frac{d}{3}$.
The time taken for each segment is $t_{1} = \frac{d/3}{v_{1}}$,$t_{2} = \frac{d/3}{v_{2}}$,and $t_{3} = \frac{d/3}{v_{3}}$.
The average velocity $v_{avg}$ is given by:
$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{d}{\frac{d}{3v_{1}} + \frac{d}{3v_{2}} + \frac{d}{3v_{3}}}$
$v_{avg} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$
Given $v_{1} = 11 \, ms^{-1}$,$v_{2} = 2v_{1} = 22 \, ms^{-1}$,and $v_{3} = 3v_{1} = 33 \, ms^{-1}$.
Substituting these values:
$v_{avg} = \frac{3}{\frac{1}{11} + \frac{1}{22} + \frac{1}{33}} = \frac{3}{\frac{6+3+2}{66}} = \frac{3 \times 66}{11} = 3 \times 6 = 18 \, ms^{-1}$.

(C) Let the total distance between $P$ and $Q$ be $3x \, km$.
The distance is divided into three equal parts,each of length $x \, km$.
For the first part,the speed is $v_1 = 10 \, km/h$. The time taken is $t_1 = \frac{x}{10} \, h$.
For the second part,the speed is $v_2 = 20 \, km/h$. The time taken is $t_2 = \frac{x}{20} \, h$.
For the third part,the speed is $v_3 = 60 \, km/h$. The time taken is $t_3 = \frac{x}{60} \, h$.
The average speed is defined as the total distance divided by the total time taken:
$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3x}{t_1 + t_2 + t_3}$
Substituting the values:
$\text{Average Speed} = \frac{3x}{\frac{x}{10} + \frac{x}{20} + \frac{x}{60}}$
Taking the common denominator $(60)$:
$\text{Average Speed} = \frac{3x}{\frac{6x + 3x + x}{60}} = \frac{3x}{\frac{10x}{60}} = \frac{3x \times 60}{10x} = 18 \, km/h$.

(B) Average speed is defined as the total distance traveled divided by the total time taken.
From the graph,the particle moves from $x=0$ to $x=10$ in the interval $t=0$ to $t=2 \, s$ (distance = $10 \, m$).
From $t=2 \, s$ to $t=4 \, s$,the particle is at rest (distance = $0 \, m$).
From $t=4 \, s$ to $t=6 \, s$,the particle moves from $x=10$ to $x=20$ (distance = $10 \, m$).
From $t=6 \, s$ to $t=8 \, s$,the particle moves from $x=20$ to $x=0$ (distance = $20 \, m$).
Total distance = $10 + 0 + 10 + 20 = 40 \, m$.
Total time = $8 \, s$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{40}{8} = 5 \, m/s$.

(B) The average velocity is defined as $\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$.
If the average velocity is $0$,then the total displacement must be $0$.
For a particle moving in a straight line to have a displacement of $0$,it must return to its starting position.
To return to the starting position,the particle must change its direction of motion at some point.
At the instant the particle changes its direction,its velocity must be $0$ (assuming continuous motion).
Therefore,the velocity of the particle must be zero at at least one instant in the interval.

(B) Let the total distance be $2d$. The first half distance $d$ is covered with speed $v_1 = 6 \,m/s$.
The remaining distance $d$ is covered in two parts,each for time $t$. Let the speeds be $v_2 = 2 \,m/s$ and $v_3 = 4 \,m/s$.
For the second half of the journey,the average speed $v'$ is the arithmetic mean of the speeds because the time intervals are equal:
$v' = \frac{v_2 + v_3}{2} = \frac{2 + 4}{2} = 3 \,m/s$.
Now,the total journey consists of two equal distances $d$ covered with speeds $v_1 = 6 \,m/s$ and $v' = 3 \,m/s$. The average speed for equal distances is given by the harmonic mean:
$v_{av} = \frac{2 v_1 v'}{v_1 + v'} = \frac{2 \times 6 \times 3}{6 + 3} = \frac{36}{9} = 4 \,m/s$.

(D) The average speed is defined as the ratio of the total distance traveled to the total time taken.
Total distance = $x + x = 2x$.
Time taken for the first half,$t_1 = \frac{x}{v_1}$.
Time taken for the second half,$t_2 = \frac{x}{v_2}$.
Total time = $t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x \left( \frac{v_1 + v_2}{v_1 v_2} \right)$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x \left( \frac{v_1 + v_2}{v_1 v_2} \right)} = \frac{2v_1 v_2}{v_1 + v_2}$.

(D) The average speed is defined as the total distance traveled divided by the total time taken.
Total distance $d_{total} = 4\,km + 4\,km = 8\,km$.
Time taken for the first part $t_1 = \frac{d_1}{v_1} = \frac{4}{3}\,h$.
Time taken for the second part $t_2 = \frac{d_2}{v_2} = \frac{4}{5}\,h$.
Total time $t_{total} = t_1 + t_2 = \frac{4}{3} + \frac{4}{5} = \frac{20 + 12}{15} = \frac{32}{15}\,h$.
Average speed $V_{av} = \frac{d_{total}}{t_{total}} = \frac{8}{32/15} = \frac{8 \times 15}{32} = \frac{15}{4} = 3.75\,km/h$.

(C) Let $AB = x$.
Given $AB = BC$,so $BC = x$.
Given $AD = 3 AB$,so $AD = 3x$.
Since $AD = AB + BC + CD$,we have $3x = x + x + CD$,which implies $CD = x$.
The total distance covered is $D_{total} = AB + BC + CD = x + x + x = 3x$.
The total time taken is $T_{total} = t_1 + t_2 + t_3 = \frac{AB}{v_1} + \frac{BC}{v_2} + \frac{CD}{v_3} = \frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}$.
The average speed is defined as $v_{avg} = \frac{D_{total}}{T_{total}} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$.
Simplifying the expression,we get $v_{avg} = \frac{3x}{x(\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3})} = \frac{3}{\frac{v_2 v_3 + v_1 v_3 + v_1 v_2}{v_1 v_2 v_3}} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$.

(C) The average velocity is defined as the total displacement divided by the total time taken.
Total distance $= x + x = 2x$.
Time taken for the first part of the journey $t_1 = \frac{x}{v_1}$.
Time taken for the second part of the journey $t_2 = \frac{x}{v_2}$.
Total time taken $T = t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x \left( \frac{1}{v_1} + \frac{1}{v_2} \right)$.
Average velocity $v = \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x \left( \frac{1}{v_1} + \frac{1}{v_2} \right)}$.
Canceling $x$ from the numerator and denominator,we get $v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.
Rearranging this gives $\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$.

(D) Let the total distance be $S$.
The vehicle travels the first half distance $(S/2)$ with speed $v_1 = v$.
The time taken for the first half is $t_1 = \frac{S/2}{v} = \frac{S}{2v}$.
The vehicle travels the remaining half distance $(S/2)$ with speed $v_2 = 2v$.
The time taken for the second half is $t_2 = \frac{S/2}{2v} = \frac{S}{4v}$.
The average speed is defined as the total distance divided by the total time:
$V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{S}{t_1 + t_2}$
$V_{avg} = \frac{S}{\frac{S}{2v} + \frac{S}{4v}} = \frac{S}{\frac{2S + S}{4v}} = \frac{S}{\frac{3S}{4v}}$
$V_{avg} = \frac{4v}{3}$

(A) The average velocity is defined as the total displacement divided by the total time taken.
$v_{\text{avg}} = \frac{x_1 + x_2}{t_1 + t_2}$
Given $x_1 = x$,$x_2 = \frac{3}{2}x$,$v_1 = 5 \ m/s$,and $v_{\text{avg}} = \frac{50}{7} \ m/s$.
The time taken for the first part is $t_1 = \frac{x}{v_1} = \frac{x}{5}$.
The time taken for the second part is $t_2 = \frac{x_2}{v_2} = \frac{3x}{2v_2}$.
Substituting these into the average velocity formula:
$\frac{50}{7} = \frac{x + \frac{3}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}$
$\frac{50}{7} = \frac{\frac{5}{2}x}{x(\frac{1}{5} + \frac{3}{2v_2})}$
$\frac{50}{7} = \frac{2.5}{\frac{1}{5} + \frac{3}{2v_2}}$
$\frac{1}{5} + \frac{3}{2v_2} = \frac{2.5 \times 7}{50} = \frac{17.5}{50} = \frac{7}{20}$
$\frac{3}{2v_2} = \frac{7}{20} - \frac{1}{5} = \frac{7-4}{20} = \frac{3}{20}$
$\frac{3}{2v_2} = \frac{3}{20} \Rightarrow 2v_2 = 20 \Rightarrow v_2 = 10 \ m/s$.

(C) The total distance is $L$. The vehicle travels the first half distance $(L/2)$ with speed $V_1 = V$ and the second half distance $(L/2)$ with speed $V_2 = \frac{V}{3}$.
Time taken for the first half,$t_1 = \frac{L/2}{V} = \frac{L}{2V}$.
Time taken for the second half,$t_2 = \frac{L/2}{V/3} = \frac{3L}{2V}$.
Total time taken,$T = t_1 + t_2 = \frac{L}{2V} + \frac{3L}{2V} = \frac{4L}{2V} = \frac{2L}{V}$.
Average speed,$V_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time}} = \frac{L}{2L/V} = \frac{V}{2}$.

(A) Let the total distance covered by the body be $2d$.
Then,the first half distance is $d$ and the second half distance is $d$.
Time taken to cover the first half distance at speed $u$ is $t_1 = \frac{d}{u}$.
Time taken to cover the second half distance at speed $v$ is $t_2 = \frac{d}{v}$.
Average speed is defined as the total distance divided by the total time taken.
Average speed $= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{t_1 + t_2} = \frac{2d}{\frac{d}{u} + \frac{d}{v}}$.
Simplifying the expression: $\frac{2d}{d(\frac{1}{u} + \frac{1}{v})} = \frac{2}{\frac{u+v}{uv}} = \frac{2uv}{u+v}$.

(B) Let the distance between $A$ and $B$ be $d$.
Time taken to travel from $A$ to $B$ is $t_{1} = \frac{d}{v_{1}} = \frac{d}{30}$.
Time taken to travel from $B$ to $A$ is $t_{2} = \frac{d}{v_{2}} = \frac{d}{20}$.
Average speed is defined as the total distance divided by the total time.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{d + d}{t_{1} + t_{2}} = \frac{2d}{\frac{d}{30} + \frac{d}{20}}$.
Average speed $= \frac{2d}{d(\frac{20 + 30}{600})} = \frac{2 \times 600}{50} = \frac{1200}{50} = 24 \ km/h$.

(C) Average speed is defined as the total distance traveled divided by the total time taken.
From the graph,we can calculate the distance traveled in each interval:
$1$. From $t = 0 \text{ s}$ to $t = 1 \text{ s}$,distance = $|2 - 1| = 1 \text{ m}$.
$2$. From $t = 1 \text{ s}$ to $t = 2 \text{ s}$,distance = $|3 - 2| = 1 \text{ m}$.
$3$. From $t = 2 \text{ s}$ to $t = 3 \text{ s}$,distance = $|3 - 3| = 0 \text{ m}$.
$4$. From $t = 3 \text{ s}$ to $t = 4 \text{ s}$,distance = $|2 - 3| = 1 \text{ m}$.
$5$. From $t = 4 \text{ s}$ to $t = 5 \text{ s}$,distance = $|3 - 2| = 1 \text{ m}$.
Total distance = $1 + 1 + 0 + 1 + 1 = 4 \text{ m}$.
Total time = $5 \text{ s}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{4 \text{ m}}{5 \text{ s}} = 0.8 \text{ m/s}$.

(C) Given,velocity $v = 6t - 3t^2$.
We know that velocity $v = \frac{dx}{dt}$,where $x$ is the displacement.
Therefore,the displacement $x$ is given by the integral of velocity with respect to time:
$x = \int_{0}^{2} v \ dt = \int_{0}^{2} (6t - 3t^2) \ dt$
$x = \left[ \frac{6t^2}{2} - \frac{3t^3}{3} \right]_{0}^{2} = \left[ 3t^2 - t^3 \right]_{0}^{2}$
$x = (3(2)^2 - (2)^3) - (3(0)^2 - (0)^3)$
$x = (12 - 8) - 0 = 4 \ m$.
The average velocity $v_{avg}$ is defined as the total displacement divided by the total time taken:
$v_{avg} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{4 \ m}{2 \ s} = 2 \ m/s$.
Thus,the average velocity is $2 \ m/s$.

(D) Let the total distance be $D$.
Distance covered with speed $v_1$ is $d_1 = 0.4D$.
Distance covered with speed $v_2$ is $d_2 = 0.6D$.
Time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{0.4D}{v_1}$.
Time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{0.6D}{v_2}$.
Average speed $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D}{t_1 + t_2}$.
$v_{avg} = \frac{D}{\frac{0.4D}{v_1} + \frac{0.6D}{v_2}} = \frac{1}{\frac{0.4}{v_1} + \frac{0.6}{v_2}}$.
$v_{avg} = \frac{1}{\frac{0.4v_2 + 0.6v_1}{v_1 v_2}} = \frac{v_1 v_2}{0.4v_2 + 0.6v_1}$.
Multiply numerator and denominator by $5$ to simplify: $v_{avg} = \frac{5 v_1 v_2}{2v_2 + 3v_1} = \frac{5 v_1 v_2}{3v_1 + 2v_2}$.

(A) Average velocity is defined as the total displacement divided by the total time interval.
From the graph,at $t = 0 \ s$,the initial position $x_i = 80 \ m$.
At $t = 10 \ s$,the final position $x_f = 60 \ m$.
The total displacement $\Delta x = x_f - x_i = 60 \ m - 80 \ m = -20 \ m$.
The total time interval $\Delta t = 10 \ s - 0 \ s = 10 \ s$.
Average velocity $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{-20 \ m}{10 \ s} = -2 \ m \ s^{-1}$.
The magnitude of the average velocity is $2 \ m \ s^{-1}$.

(C) Let the total distance be $d$.
Time taken for the first half distance $(d/2)$ is $t_1 = \frac{d/2}{V_1} = \frac{d}{2V_1}$.
Time taken for the second half distance $(d/2)$ is $t_2 = \frac{d/2}{V_2} = \frac{d}{2V_2}$.
Average velocity is defined as the total distance divided by the total time taken.
Average velocity $= \frac{\text{Total distance}}{\text{Total time}} = \frac{d}{t_1 + t_2}$.
Substituting the values of $t_1$ and $t_2$:
Average velocity $= \frac{d}{\frac{d}{2V_1} + \frac{d}{2V_2}} = \frac{d}{\frac{d}{2} \left( \frac{1}{V_1} + \frac{1}{V_2} \right)} = \frac{1}{\frac{1}{2} \left( \frac{V_1 + V_2}{V_1 V_2} \right)} = \frac{2V_1 V_2}{V_1 + V_2}$.
Note that $\frac{2V_1 V_2}{V_1 + V_2}$ is equivalent to $\frac{2}{\frac{1}{V_1} + \frac{1}{V_2}}$,which matches option $C$.

(A) Given that, speed of train from city-$A$ to city-$B$ is $v_1 = 18 \,ms^{-1}$.
Speed of train from city-$B$ to city-$A$ is $v_2 = 36 \,ms^{-1}$.
Since the distance travelled from $A$ to $B$ and $B$ to $A$ is the same, we use the formula for average speed over equal distances:
$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$
Substituting the values:
$v_{av} = \frac{2 \times 18 \times 36}{18 + 36}$
$v_{av} = \frac{2 \times 18 \times 36}{54}$
$v_{av} = \frac{1296}{54} = 24 \,ms^{-1}$.

(B) Let the total distance be $2d$. The first half distance is $d$ and the second half distance is $d$.
For the first half distance $(d)$: Velocity $v_1 = 7 \,m/s$. Time taken $t_1 = \frac{d}{v_1} = \frac{d}{7}$.
For the second half distance $(d)$: Let the total time taken be $t_2$. The first half of this time $(t_2/2)$ is travelled with $v_2 = 14 \,m/s$ and the second half $(t_2/2)$ with $v_3 = 21 \,m/s$.
Distance $d = (v_2 \times \frac{t_2}{2}) + (v_3 \times \frac{t_2}{2}) = (14 \times \frac{t_2}{2}) + (21 \times \frac{t_2}{2}) = 7t_2 + 10.5t_2 = 17.5t_2$.
So, $t_2 = \frac{d}{17.5} = \frac{d}{35/2} = \frac{2d}{35}$.
Total distance = $2d$.
Total time $T = t_1 + t_2 = \frac{d}{7} + \frac{2d}{35} = \frac{5d + 2d}{35} = \frac{7d}{35} = \frac{d}{5}$.
Average velocity $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/5} = 10 \,m/s$.

(A) Let the car cover a distance $s$ from point $A$ to $B$ and then return to point $A$.
The average speed is defined as the total distance covered divided by the total time taken.
Time taken from $A$ to $B$ is $t_1 = \frac{s}{60}$.
Time taken from $B$ to $A$ is $t_2 = \frac{s}{V}$.
The total distance is $s + s = 2s$.
The total time is $t_1 + t_2 = \frac{s}{60} + \frac{s}{V}$.
Given the average speed is $48 \ km \ h^{-1}$,we have:
$48 = \frac{2s}{\frac{s}{60} + \frac{s}{V}}$
Dividing both sides by $s$ (assuming $s \neq 0$):
$48 = \frac{2}{\frac{1}{60} + \frac{1}{V}}$
$\frac{1}{60} + \frac{1}{V} = \frac{2}{48} = \frac{1}{24}$
$\frac{1}{V} = \frac{1}{24} - \frac{1}{60}$
$\frac{1}{V} = \frac{5 - 2}{120} = \frac{3}{120} = \frac{1}{40}$
Therefore,$V = 40 \ km \ h^{-1}$.

(B) The average speed is defined as the total distance divided by the total time taken.
Let the total distance be $L$.
The distance for the first part is $d_1 = \frac{L}{3}$ and the speed is $v_1$.
The time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{L/3}{v_1} = \frac{L}{3 v_1}$.
The distance for the second part is $d_2 = \frac{2L}{3}$ and the speed is $v_2$.
The time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{2L/3}{v_2} = \frac{2L}{3 v_2}$.
The total time taken is $T = t_1 + t_2 = \frac{L}{3 v_1} + \frac{2L}{3 v_2} = \frac{L}{3} \left( \frac{1}{v_1} + \frac{2}{v_2} \right) = \frac{L}{3} \left( \frac{v_2 + 2 v_1}{v_1 v_2} \right)$.
The average speed $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{L}{\frac{L}{3} \left( \frac{2 v_1 + v_2}{v_1 v_2} \right)} = \frac{3 v_1 v_2}{2 v_1 + v_2}$.

(A) The average velocity $v_{avg}$ is defined as the total change in displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$.
Given $x(t) = \alpha t + \beta t^3$ with $\alpha = 2.0 \ m \ s^{-1}$ and $\beta = 0.01 \ m \ s^{-3}$.
At $t_1 = 2.00 \ s$: $x(2) = 2.0(2) + 0.01(2)^3 = 4.0 + 0.01(8) = 4.08 \ m$.
At $t_2 = 4.00 \ s$: $x(4) = 2.0(4) + 0.01(4)^3 = 8.0 + 0.01(64) = 8.64 \ m$.
Now,calculate the average velocity: $v_{avg} = \frac{8.64 - 4.08}{4.00 - 2.00} = \frac{4.56}{2} = 2.28 \ m \ s^{-1}$.

(B) The position is given by $x(t) = \alpha + \beta t^2$.
Given $\alpha = 8 \ m$,the equation becomes $x(t) = 8 + \beta t^2$.
The average velocity $v_{avg}$ between $t_1 = 2 \ s$ and $t_2 = 4 \ s$ is defined as $v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$.
Calculate the position at $t_1 = 2 \ s$: $x(2) = 8 + \beta(2)^2 = 8 + 4\beta$.
Calculate the position at $t_2 = 4 \ s$: $x(4) = 8 + \beta(4)^2 = 8 + 16\beta$.
The change in position is $\Delta x = x(4) - x(2) = (8 + 16\beta) - (8 + 4\beta) = 12\beta$.
The time interval is $\Delta t = 4 - 2 = 2 \ s$.
Given $v_{avg} = 12 \ m/s$,we have $12 = \frac{12\beta}{2}$.
$12 = 6\beta$,which implies $\beta = 2 \ m/s^2$.

(C) Given that the distances travelled by the body are in the ratio $1: 2$.
Let the distance travelled with speed $v_1$ be $s$. Then,the distance travelled with speed $v_2$ will be $2s$.
Average speed is defined as the total distance travelled divided by the total time taken.
Total distance = $s + 2s = 3s$.
Time taken for the first part,$t_1 = \frac{s}{v_1}$.
Time taken for the second part,$t_2 = \frac{2s}{v_2}$.
Total time taken,$T = t_1 + t_2 = \frac{s}{v_1} + \frac{2s}{v_2} = s \left( \frac{1}{v_1} + \frac{2}{v_2} \right) = s \left( \frac{v_2 + 2v_1}{v_1 v_2} \right)$.
Average speed,$v_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time}} = \frac{3s}{s \left( \frac{v_2 + 2v_1}{v_1 v_2} \right)} = \frac{3 v_1 v_2}{v_2 + 2v_1}$.

(B) Time taken to reach the market: $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$.
Time taken to return to the house: $t_2 = \frac{\text{distance}}{\text{speed}} = \frac{2.5 \,km}{7.5 \,km/h} = \frac{1}{3} \,h = 20 \,min$.
Total time taken for the round trip is $30 \,min + 20 \,min = 50 \,min$.
Total distance covered = $2.5 \,km + 2.5 \,km = 5 \,km = 5000 \,m$.
Total time in seconds = $50 \,min \times 60 \,s/min = 3000 \,s$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{5000 \,m}{3000 \,s} = \frac{5}{3} \,m/s$.