Average Velocity and Average Speed Questions in English

(D) Let the total distance be $2d$.
The person travels distance $d$ with velocity $v_1$ and the remaining distance $d$ with velocity $v_2$.
Time taken for the first half is $t_1 = \frac{d}{v_1}$.
Time taken for the second half is $t_2 = \frac{d}{v_2}$.
Average velocity $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2}$.
Substituting the values: $v_{avg} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}} = \frac{2d}{d(\frac{v_1 + v_2}{v_1 v_2})} = \frac{2v_1 v_2}{v_1 + v_2}$.
Thus,the correct option is $D$.

(B) The formula for the average speed when a body travels a distance $d$ at speed $v_1$ and returns the same distance $d$ at speed $v_2$ is given by:
$\text{Average speed} = \frac{2v_1v_2}{v_1 + v_2}$
Given $v_1 = 20\,km/hr$ and $v_2 = 30\,km/hr$.
Substituting the values:
$\text{Average speed} = \frac{2 \times 20 \times 30}{20 + 30}$
$\text{Average speed} = \frac{1200}{50}$
$\text{Average speed} = 24\,km/hr$
Therefore,the correct option is $(b)$.

(B) The average speed for a round trip where the distance for both legs is the same is given by the formula: $v_{avg} = \frac{2v_1v_2}{v_1 + v_2}$.
Here,$v_1 = 2.5\, km/h$ and $v_2 = 4\, km/h$.
Substituting the values into the formula:
$v_{avg} = \frac{2 \times 2.5 \times 4}{2.5 + 4}$
$v_{avg} = \frac{20}{6.5}$
$v_{avg} = \frac{200}{65} = \frac{40}{13}\, km/h$.
Thus,the correct option is $B$.

(C) When a body covers two equal distances with different speeds $v_1$ and $v_2$,the average speed is given by the harmonic mean formula: $\text{Average Speed} = \frac{2v_1v_2}{v_1 + v_2}$.
Given $v_1 = 30 \, km/hr$ and $v_2 = 50 \, km/hr$.
Substituting the values: $\text{Average Speed} = \frac{2 \times 30 \times 50}{30 + 50}$.
$\text{Average Speed} = \frac{3000}{80}$.
$\text{Average Speed} = \frac{300}{8} = 37.5 \, km/hr$.

(D) Let the total distance be $x$.
Time taken to cover the first one-third distance $(x/3)$ at $v_1 = 20 \, km/hr$ is $t_1 = \frac{x/3}{20} = \frac{x}{60} \, hr$.
Time taken to cover the remaining two-thirds distance $(2x/3)$ at $v_2 = 60 \, km/hr$ is $t_2 = \frac{2x/3}{60} = \frac{2x}{180} = \frac{x}{90} \, hr$.
Average speed is defined as the total distance divided by the total time: $v_{avg} = \frac{x}{t_1 + t_2}$.
$v_{avg} = \frac{x}{\frac{x}{60} + \frac{x}{90}} = \frac{1}{\frac{1}{60} + \frac{1}{90}}$.
Finding a common denominator: $\frac{1}{60} + \frac{1}{90} = \frac{3+2}{180} = \frac{5}{180} = \frac{1}{36}$.
Therefore,$v_{avg} = \frac{1}{1/36} = 36 \, km/hr$.

(A) When a body travels for equal intervals of time with different speeds $v_1$ and $v_2$,the average speed is given by the arithmetic mean of the speeds.
Average speed $v_{avg} = \frac{v_1 + v_2}{2}$.
Given $v_1 = 80 \, km/h$ and $v_2 = 40 \, km/h$.
$v_{avg} = \frac{80 + 40}{2} = \frac{120}{2} = 60 \, km/h$.
Thus,the average speed of the car is $60 \, km/h$.

(B) The distance travelled by the train in the first $1\, h$ is $d_1 = 60\, km/h \times 1\, h = 60\, km$.
The distance travelled by the train in the next $0.5\, h$ is $d_2 = 40\, km/h \times 0.5\, h = 20\, km$.
The total distance travelled is $D = d_1 + d_2 = 60\, km + 20\, km = 80\, km$.
The total time taken is $T = 1\, h + 0.5\, h = 1.5\, h = 3/2\, h$.
The average speed is given by $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{80}{1.5} = \frac{80}{3/2} = \frac{160}{3} \approx 53.33\, km/h$.

(D) Average velocity is defined as the ratio of total displacement to total time taken.
Displacement is the shortest distance between the initial and final positions.
For a particle moving along a semicircle of radius $r = 10\,m$,the displacement is equal to the diameter of the circle.
Displacement $= 2r = 2 \times 10\,m = 20\,m$.
Total time taken $= 5\,s$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{20\,m}{5\,s} = 4\,m/s$.
Therefore,the correct option is $D$.

(D) Time taken to reach the market: $t_1 = \frac{d}{v_1} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$.
Since the total time interval is $40 \,min$,the time spent walking back is $t_2 = 40 \,min - 30 \,min = 10 \,min = \frac{10}{60} \,h = \frac{1}{6} \,h$.
Distance covered while returning: $d_2 = v_2 \times t_2 = 7.5 \,km/h \times \frac{1}{6} \,h = 1.25 \,km$.
Total distance covered: $D = 2.5 \,km + 1.25 \,km = 3.75 \,km$.
Total time taken: $T = 40 \,min = \frac{40}{60} \,h = \frac{2}{3} \,h$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{3.75 \,km}{(2/3) \,h} = 3.75 \times 1.5 = 5.625 \,km/h = \frac{45}{8} \,km/h$.

(B) The average velocity is defined as the ratio of total displacement to total time,while average speed is the ratio of total distance to total time.
Mathematically,$\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}}$ and $\text{Average speed} = \frac{\text{Distance}}{\text{Time}}$.
Therefore,the ratio is $\frac{|\text{Average velocity}|}{|\text{Average speed}|} = \frac{|\text{Displacement}|}{\text{Distance}}$.
Since the magnitude of displacement is always less than or equal to the distance traveled $(|\text{Displacement}| \le \text{Distance})$,the ratio must be $\le 1$.
Thus,the ratio is unity or less.

(B) Let the total time taken be $T$. The person travels for the first half time $t = \frac{T}{2}$ with velocity $v_1$,so the distance covered is $s_1 = v_1 \times \frac{T}{2}$.
The person travels for the next half time $t = \frac{T}{2}$ with velocity $v_2$,so the distance covered is $s_2 = v_2 \times \frac{T}{2}$.
The mean velocity $V$ is defined as the total displacement divided by the total time.
$V = \frac{s_1 + s_2}{T} = \frac{v_1(T/2) + v_2(T/2)}{T} = \frac{(v_1 + v_2)(T/2)}{T} = \frac{v_1 + v_2}{2}$.

(B) The average velocity is defined as the total displacement divided by the total time taken.
Total displacement $S = S_1 + S_2 + S_3 = (v_1 \times t_1) + (v_2 \times t_2) + (v_3 \times t_3)$.
Given $v_1 = 3 \,m/s$,$v_2 = 4 \,m/s$,$v_3 = 5 \,m/s$ and $t_1 = t_2 = t_3 = 20 \,s$.
$S = (3 \times 20) + (4 \times 20) + (5 \times 20) = 60 + 80 + 100 = 240 \,m$.
Total time $T = 20 + 20 + 20 = 60 \,s$.
Average velocity $v_{avg} = \frac{S}{T} = \frac{240}{60} = 4 \,m/s$.
Since the time intervals are equal,the average velocity is simply the arithmetic mean of the velocities: $v_{avg} = \frac{v_1 + v_2 + v_3}{3} = \frac{3 + 4 + 5}{3} = 4 \,m/s$.

(C) Let the total distance be $2d$. The car travels the first half distance $d$ with velocity $v_1 = 40 \, km/h$ and the second half distance $d$ with velocity $v_2 = 60 \, km/h$.
Time taken for the first half is $t_1 = \frac{d}{v_1} = \frac{d}{40}$.
Time taken for the second half is $t_2 = \frac{d}{v_2} = \frac{d}{60}$.
Average velocity $v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2} = \frac{2d}{\frac{d}{40} + \frac{d}{60}}$.
$v_{av} = \frac{2}{\frac{1}{40} + \frac{1}{60}} = \frac{2}{\frac{3+2}{120}} = \frac{2 \times 120}{5} = \frac{240}{5} = 48 \, km/h$.

(A) Let the total distance be $x$. The first half distance is $x/2$ covered with speed $v_1 = 3 \, m/s$. The time taken is $t_1 = \frac{x/2}{3} = \frac{x}{6}$.
The second half distance $x/2$ is covered in two equal time intervals,say $t_2$ each. Let the speeds be $v_2 = 4.5 \, m/s$ and $v_3 = 7.5 \, m/s$.
Distance covered in the second half is $(v_2 \cdot t_2) + (v_3 \cdot t_2) = x/2$.
$(4.5 + 7.5) t_2 = x/2 \Rightarrow 12 t_2 = x/2 \Rightarrow t_2 = x/24$.
The total time taken for the second half is $2 t_2 = 2(x/24) = x/12$.
Total time $T = t_1 + 2 t_2 = \frac{x}{6} + \frac{x}{12} = \frac{2x + x}{12} = \frac{3x}{12} = \frac{x}{4}$.
Average speed $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{x/4} = 4 \, m/s$.

(B) The average velocity is defined as the total displacement divided by the total time taken.
Displacement is the shortest distance between the initial point $A$ and the final point $B$.
Since the particle moves along a semicircle of radius $r = 1.0 \, m$,the displacement is equal to the diameter of the semicircle.
Displacement $= 2 \times r = 2 \times 1.0 \, m = 2.0 \, m$.
The time taken is $1.0 \, s$.
Therefore,the magnitude of the average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{2.0 \, m}{1.0 \, s} = 2.0 \, m/s$.

(D) Let the total distance be $x$.
Time taken to cover the first part of the distance $(d_1 = \frac{2}{5}x)$ with speed $v_1$ is $t_1 = \frac{d_1}{v_1} = \frac{2x}{5v_1}$.
Time taken to cover the second part of the distance $(d_2 = \frac{3}{5}x)$ with speed $v_2$ is $t_2 = \frac{d_2}{v_2} = \frac{3x}{5v_2}$.
Average speed is defined as the total distance divided by the total time taken:
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2}$
Substituting the values of $t_1$ and $t_2$:
$\text{Average speed} = \frac{x}{\frac{2x}{5v_1} + \frac{3x}{5v_2}}$
$\text{Average speed} = \frac{x}{\frac{2xv_2 + 3xv_1}{5v_1 v_2}}$
$\text{Average speed} = \frac{5v_1 v_2}{3v_1 + 2v_2}$

(B) Let the total distance be $d = 200 \; m$. The first half distance is $d_1 = 100 \; m$ and the second half distance is $d_2 = 100 \; m$.
The time taken for the first half is $t_1 = \frac{d_1}{v_1} = \frac{100}{40} \; h$.
The time taken for the second half is $t_2 = \frac{d_2}{v} = \frac{100}{v} \; h$.
The average speed $v_{avg}$ is given by the formula: $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d}{t_1 + t_2}$.
Substituting the given values: $48 = \frac{200}{\frac{100}{40} + \frac{100}{v}}$.
Dividing the numerator and denominator by $100$: $48 = \frac{2}{\frac{1}{40} + \frac{1}{v}}$.
Rearranging the terms: $\frac{1}{40} + \frac{1}{v} = \frac{2}{48} = \frac{1}{24}$.
Solving for $v$: $\frac{1}{v} = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{2}{120} = \frac{1}{60}$.
Therefore,$v = 60 \; km/h$.

(C) Let the total distance be $x$.
Time taken for the first part of the journey $(d_1 = x/3)$ at speed $v_1 = 20 \, km/hr$ is $t_1 = \frac{x/3}{20} = \frac{x}{60} \, hr$.
Time taken for the remaining part of the journey $(d_2 = 2x/3)$ at speed $v_2 = 60 \, km/hr$ is $t_2 = \frac{2x/3}{60} = \frac{2x}{180} = \frac{x}{90} \, hr$.
Total time $T = t_1 + t_2 = \frac{x}{60} + \frac{x}{90} = \frac{3x + 2x}{180} = \frac{5x}{180} = \frac{x}{36} \, hr$.
Average speed $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{x/36} = 36 \, km/hr$.

(C) Average speed is defined as the ratio of total distance traveled to the total time taken.
Let the distance between $X$ and $Y$ be $d$.
Time taken to travel from $X$ to $Y$ is $t_1 = \frac{d}{v_1}$.
Time taken to travel from $Y$ to $X$ is $t_2 = \frac{d}{v_2}$.
Total distance $= d + d = 2d$.
Total time $= t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}$.
Average speed $\bar v = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}}$.
Canceling $d$ from the numerator and denominator,we get $\bar v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.
Rearranging this gives $\frac{2}{\bar v} = \frac{1}{v_1} + \frac{1}{v_2}$.

(C) Let the total distance be $d$. The particle covers the first half distance $(d/2)$ with speed $v_1$ and the second half distance $(d/2)$ with speed $v_2$.
Time taken for the first half,$t_1 = \frac{d/2}{v_1} = \frac{d}{2v_1}$.
Time taken for the second half,$t_2 = \frac{d/2}{v_2} = \frac{d}{2v_2}$.
Average speed is defined as the total distance divided by the total time taken:
$v_{av} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d}{t_1 + t_2}$.
Substituting the values of $t_1$ and $t_2$:
$v_{av} = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{d}{\frac{d}{2} (\frac{1}{v_1} + \frac{1}{v_2})} = \frac{1}{\frac{1}{2} (\frac{v_1 + v_2}{v_1 v_2})}$.
$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.

(D) Average velocity is defined as the total displacement divided by the total time taken.
Since the car returns to its starting point,the final position is the same as the initial position.
Therefore,the total displacement is $0$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{0}{2 + 3} = 0 \text{ km/h}$.
Thus,the correct option is $D$.

(C) The average speed is defined as the total distance divided by the total time taken.
Let the total distance be $S$.
The distance for each segment is $S/3$.
The time taken for the first segment is $t_1 = \frac{S/3}{V} = \frac{S}{3V}$.
The time taken for the second segment is $t_2 = \frac{S/3}{2V} = \frac{S}{6V}$.
The time taken for the third segment is $t_3 = \frac{S/3}{3V} = \frac{S}{9V}$.
Total time $T = t_1 + t_2 + t_3 = \frac{S}{3V} + \frac{S}{6V} + \frac{S}{9V}$.
Taking the common denominator as $18V$,we get $T = \frac{6S + 3S + 2S}{18V} = \frac{11S}{18V}$.
Average speed $V_{av} = \frac{S}{T} = \frac{S}{11S / 18V} = \frac{18V}{11}$.

(D) Let the total distance be $x$.
The distance covered in each part is $x/3$.
The time taken for the first part is $t_1 = \frac{x/3}{v_1} = \frac{x}{3v_1}$.
The time taken for the second part is $t_2 = \frac{x/3}{v_2} = \frac{x}{3v_2}$.
The time taken for the third part is $t_3 = \frac{x/3}{v_3} = \frac{x}{3v_3}$.
Average speed $v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2 + t_3}$.
$v_{av} = \frac{x}{\frac{x}{3v_1} + \frac{x}{3v_2} + \frac{x}{3v_3}} = \frac{1}{\frac{1}{3v_1} + \frac{1}{3v_2} + \frac{1}{3v_3}}$.
$v_{av} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3}{\frac{v_2 v_3 + v_1 v_3 + v_1 v_2}{v_1 v_2 v_3}}$.
$v_{av} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$.

(A) Let the side length of the regular hexagon be $a$. The particle moves with constant speed $v$.
For motion from $A$ to $F$,the displacement is the vector $\vec{AF}$. Since $ABCDEF$ is a regular hexagon,the distance $AF$ is equal to the side length $a$.
The time taken to travel from $A$ to $F$ along the path $A \to B \to C \to D \to E \to F$ is $t = \frac{5a}{v}$.
The magnitude of average velocity is given by $|\vec{v}_{av}| = \frac{|\text{displacement}|}{\text{time taken}}$.
$|\vec{v}_{av}| = \frac{a}{5a/v} = \frac{v}{5}$.

(D) Average velocity is defined as the total displacement divided by the total time,or the vector sum of individual velocities divided by the number of particles.
Speed is a scalar quantity representing the magnitude of velocity,while velocity is a vector quantity having both magnitude and direction.
The question provides only the speeds (magnitudes) of the five particles but does not provide any information regarding the directions of their motion.
Since the direction of each particle is unknown,the vector sum of the velocities cannot be determined.
Therefore,the average velocity cannot be calculated with the given information.

(A) Let the total distance be $3x$. The distance for each segment is $x$.
The time taken for the first segment is $t_1 = \frac{x}{10}$.
The time taken for the second segment is $t_2 = \frac{x}{20}$.
The time taken for the third segment is $t_3 = \frac{x}{60}$.
The average speed $v_{avg}$ is given by the total distance divided by the total time:
$v_{avg} = \frac{3x}{t_1 + t_2 + t_3} = \frac{3x}{\frac{x}{10} + \frac{x}{20} + \frac{x}{60}}$.
Factoring out $x$:
$v_{avg} = \frac{3}{\frac{1}{10} + \frac{1}{20} + \frac{1}{60}} = \frac{3}{\frac{6+3+1}{60}} = \frac{3}{\frac{10}{60}} = \frac{3 \times 60}{10} = 18\; km/h$.

(B) The average velocity is defined as the total displacement divided by the total time taken.
First,calculate the displacement in the east direction:
$\vec{d}_1 = 15 \, m/s \times 2 \, s \times \hat{i} = 30 \hat{i} \, m$
Next,calculate the displacement in the north direction:
$\vec{d}_2 = 5 \, m/s \times 8 \, s \times \hat{j} = 40 \hat{j} \, m$
Total displacement $\vec{d} = \vec{d}_1 + \vec{d}_2 = 30 \hat{i} + 40 \hat{j} \, m$
Total time $t = 2 \, s + 8 \, s = 10 \, s$
Average velocity $\vec{v}_{av} = \frac{\vec{d}}{t} = \frac{30 \hat{i} + 40 \hat{j}}{10} = 3 \hat{i} + 4 \hat{j} \, m/s$
The magnitude of the average velocity is $|\vec{v}_{av}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \, m/s$.
The direction is given by $\tan \theta = \frac{v_y}{v_x} = \frac{4}{3}$,which implies $\theta = 53^\circ$ north of east,or $37^\circ$ east of north.
Thus,the average velocity is $5 \, m/s$ due $N-37^\circ-E$.

(C) The total distance is $S$. The distance is divided into three equal parts,each of length $S/3$.
Let the speeds in these parts be $v_1 = V$,$v_2 = 2V$,and $v_3 = 3V$.
The time taken for each part is $t_1 = \frac{S/3}{V}$,$t_2 = \frac{S/3}{2V}$,and $t_3 = \frac{S/3}{3V}$.
Average speed $V_{av} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{S}{t_1 + t_2 + t_3}$.
$V_{av} = \frac{S}{\frac{S}{3V} + \frac{S}{6V} + \frac{S}{9V}} = \frac{1}{\frac{1}{3V} + \frac{1}{6V} + \frac{1}{9V}}$.
Taking the common denominator $18V$: $V_{av} = \frac{1}{\frac{6+3+2}{18V}} = \frac{18V}{11}$.

(C) Let the distance between the starting point and the turning point be $D$.
Time taken to travel the distance $D$ with speed $v_1 = 30\, m/s$ is $T_1 = D / 30$.
Time taken to return the same distance $D$ with speed $v_2 = 20\, m/s$ is $T_2 = D / 20$.
The total distance traveled is $D + D = 2D$.
The total time taken is $T_1 + T_2 = D/30 + D/20 = (2D + 3D) / 60 = 5D / 60 = D / 12$.
The average speed is defined as the total distance divided by the total time:
Average Speed $= (2D) / (D / 12) = 2 \times 12 = 24\, m/s$.

(A) The average velocity $v_{avg}$ is defined as the total displacement divided by the total time interval.
$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt$
Given $v(t) = t - t^3$,$t_1 = 0 \ s$,and $t_2 = 2 \ s$.
$v_{avg} = \frac{1}{2 - 0} \int_{0}^{2} (t - t^3) \, dt$
$v_{avg} = \frac{1}{2} \left[ \frac{t^2}{2} - \frac{t^4}{4} \right]_{0}^{2}$
$v_{avg} = \frac{1}{2} \left[ (\frac{2^2}{2} - \frac{2^4}{4}) - (0 - 0) \right]$
$v_{avg} = \frac{1}{2} \left[ (\frac{4}{2} - \frac{16}{4}) \right]$
$v_{avg} = \frac{1}{2} [2 - 4] = \frac{1}{2} [-2] = -1 \ m/s$.
Note: Since the result is $-1 \ m/s$ and it is not among the options,the question or options may be flawed. However,based on the calculation,the value is $-1$.

(B) The distance covered in a specific time interval is given by the product of the average speed during that interval and the duration of the interval.
Distance in the first second $(S_{1})$:
$S_{1} = v_{av1} \times t_{1} = 5 \, ms^{-1} \times 1 \, s = 5 \, m$
Distance in the second second $(S_{2})$:
$S_{2} = v_{av2} \times t_{2} = 10 \, ms^{-1} \times 1 \, s = 10 \, m$
Distance in the third second $(S_{3})$:
$S_{3} = v_{av3} \times t_{3} = 15 \, ms^{-1} \times 1 \, s = 15 \, m$
Total distance covered $(S_{T})$:
$S_{T} = S_{1} + S_{2} + S_{3} = 5 + 10 + 15 = 30 \, m$

(C) Let the total distance be $d$. The path is divided into three equal parts,each of distance $d/3$.
Time taken for the first part: $t_1 = \frac{d/3}{v} = \frac{d}{3v}$.
Time taken for the second part: $t_2 = \frac{d/3}{2v} = \frac{d}{6v}$.
Time taken for the third part: $t_3 = \frac{d/3}{3v} = \frac{d}{9v}$.
Total time $T = t_1 + t_2 + t_3 = \frac{d}{3v} + \frac{d}{6v} + \frac{d}{9v}$.
Taking the common denominator as $18v$: $T = \frac{6d + 3d + 2d}{18v} = \frac{11d}{18v}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{d}{11d / 18v} = \frac{18v}{11}$.

(A) Let the total distance be $3d$. The distance for each part is $d$.
The time taken for the first part is $t_1 = \frac{d}{20}$.
The time taken for the second part is $t_2 = \frac{d}{30}$.
The time taken for the third part is $t_3 = \frac{d}{40}$.
The average velocity $v_{avg}$ is given by the total distance divided by the total time:
$v_{avg} = \frac{3d}{t_1 + t_2 + t_3} = \frac{3d}{\frac{d}{20} + \frac{d}{30} + \frac{d}{40}}$
$v_{avg} = \frac{3}{\frac{1}{20} + \frac{1}{30} + \frac{1}{40}}$
Taking the least common multiple of $20, 30, 40$,which is $120$:
$v_{avg} = \frac{3}{\frac{6 + 4 + 3}{120}} = \frac{3 \times 120}{13} = \frac{360}{13}$
$v_{avg} \approx 27.69 \, m/s \approx 28 \, m/s$.

(D) The average velocity is defined as the change in position divided by the change in time: $v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$.
Given $x(t) = a + b t^2$,we calculate the positions at $t_1 = 2.0 \; s$ and $t_2 = 4.0 \; s$.
$x(2.0) = a + b(2.0)^2 = a + 4b$.
$x(4.0) = a + b(4.0)^2 = a + 16b$.
Now,substitute these into the average velocity formula:
$v_{avg} = \frac{(a + 16b) - (a + 4b)}{4.0 - 2.0} = \frac{12b}{2.0} = 6b$.
Given $b = 2.5 \; m s^{-2}$,we have:
$v_{avg} = 6 \times 2.5 = 15 \; m s^{-1}$.

(N/A) The magnitude of displacement over an interval of time is the shortest distance (a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual distance covered by the particle in a given interval of time.
For example, consider a particle moving from point $A$ to point $B$ and then returning to point $C$, taking a total time $t$. As shown in the figure, the magnitude of displacement is $AC$, whereas the total path length is $AB + BC$.
Note that the magnitude of displacement can never be greater than the total path length. However, in cases where the particle moves in a single direction without turning back, both quantities are equal.
$(b)$ Magnitude of average velocity = $\frac{\text{Magnitude of displacement}}{\text{Time interval}}$
For the given particle, Average velocity = $\frac{AC}{t}$.
Average speed = $\frac{\text{Total path length}}{\text{Time interval}} = \frac{AB + BC}{t}$.
Since $(AB + BC) > AC$, the average speed is greater than the magnitude of average velocity. The two quantities are equal if the particle continues to move along a straight line in the same direction.

(D) Average velocity is defined as the total displacement divided by the total time taken.
Displacement is the shortest distance between the initial and final positions.
In this case,the man starts from his home and returns to his home after visiting the market.
Since the initial position (home) and the final position (home) are the same,the total displacement is $0 \; km$.
Therefore,Average velocity $= \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0 \; km}{\text{Total Time}} = 0 \; m/s$.

(A) The time taken to reach the market is $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{2.5 \; km}{5 \; km \; h^{-1}} = 0.5 \; h = 30 \; min$.
Since the time interval requested is $0$ to $30 \; min$,we only consider the motion from home to the market.
During this interval,the total distance covered is $2.5 \; km$.
The total time taken is $30 \; min = 0.5 \; h$.
Average speed is defined as $\frac{\text{Total distance}}{\text{Total time}}$.
Average speed $= \frac{2.5 \; km}{0.5 \; h} = 5 \; km \; h^{-1}$.

(B) Time taken to reach the market: $t_{1} = \frac{2.5 \; km}{5 \; km/h} = 0.5 \; h = 30 \; min$.
Time taken to return home: $t_{2} = \frac{2.5 \; km}{7.5 \; km/h} = \frac{1}{3} \; h \approx 20 \; min$.
Total time taken for the round trip: $t_{total} = 30 \; min + 20 \; min = 50 \; min = \frac{50}{60} \; h = \frac{5}{6} \; h$.
Total distance traveled: $d_{total} = 2.5 \; km + 2.5 \; km = 5 \; km$.
Average speed is defined as the total distance traveled divided by the total time taken.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{5 \; km}{5/6 \; h} = 5 \times \frac{6}{5} \; km/h = 6 \; km/h$.

(N/A) When an object is in motion,its position changes with time. The rate at which it changes its position can be found in two ways.
If we only consider the time rate of change of distance,it is speed. If we consider the time rate of change of position with direction,it is velocity.
Speed: The distance covered by an object in unit time is called speed.
Average speed: The ratio of the total distance covered during a journey to the total time taken is called average speed.
Its $SI$ unit is $m s^{-1}$ and it is a scalar quantity. Hence,its value is always positive.
Velocity: The displacement covered in unit time is called velocity. It is a vector quantity.
Average velocity is defined as the change in position or displacement $(\Delta x)$ divided by the time interval $(\Delta t)$ in which the displacement occurs:
$\bar{v} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}} = \frac{\Delta x}{\Delta t}$
where $x_{2}$ and $x_{1}$ are the positions of the object at times $t_{2}$ and $t_{1}$ respectively. The $SI$ unit for velocity is $m s^{-1}$,although $km h^{-1}$ is used in many everyday applications.
For motion in a straight line,the directional aspect of the vector can be represented by '$+$' and '$-$' signs,and we do not have to use vector notation for velocity.
The magnitude of average velocity can be positive,negative,or zero.
Average speed is always greater than or equal to the magnitude of average velocity. For uniform motion,velocity is equal to average velocity at every moment.
The figure shows the $x-t$ graph for the motion of a car,where the portion between $t = 5 \ s$ and $t = 7 \ s$ is highlighted to calculate average velocity.

(C) Average velocity is defined as the ratio of total displacement to total time taken $(v_{avg} = \frac{\Delta x}{\Delta t})$.
$1$. It does not provide information about the actual distance covered by the object,as distance depends on the path taken,whereas displacement only depends on the initial and final positions.
$2$. It does not describe the specific path of motion taken by the object between the initial and final points.
$3$. It does not provide information about the instantaneous velocities of the object at different positions along the path.

(N/A) Consider a car moving along a straight line as shown in the figure. Let the car move from $O$ to $P$ in $18 \, s$ and then return from $P$ to $Q$ in $6 \, s$.
Case $1$: Motion from $O$ to $P$
Total distance = $OP = 360 \, m$
Total displacement = $360 \, m - 0 \, m = 360 \, m$
Time taken = $18 \, s$
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{360}{18} = 20 \, ms^{-1}$
Average velocity = $\frac{\text{Total displacement}}{\text{Total time}} = \frac{360}{18} = 20 \, ms^{-1}$
Here, the magnitude of average velocity is equal to the average speed.
Case $2$: Motion from $O$ to $P$ and then to $Q$
Total distance = $OP + PQ = 360 + (360 - 240) = 360 + 120 = 480 \, m$
Total displacement = $OQ = 240 \, m - 0 \, m = 240 \, m$
Total time = $18 \, s + 6 \, s = 24 \, s$
Average speed = $\frac{480}{24} = 20 \, ms^{-1}$
Average velocity = $\frac{240}{24} = 10 \, ms^{-1}$
Here, the magnitude of average velocity $(10 \, ms^{-1})$ is not equal to the average speed $(20 \, ms^{-1})$.
Thus, the statement is not always correct and not always incorrect.

(N/A)

Average Speed	Average Velocity
$(1)$ The ratio of total path length covered to the total time taken is called average speed.	$(1)$ The ratio of total displacement to the total time taken is called average velocity.
$(2)$ It is a scalar quantity.	$(2)$ It is a vector quantity.
$(3)$ It is always positive for a moving object.	$(3)$ It may be positive,negative,or zero depending on the direction of displacement.
$(4)$ Average speed $\geq$ Magnitude of average velocity.	$(4)$ Average velocity $\leq$ Average speed.

(N/A) Speed: The distance covered by an object in unit time is called speed. It is defined as the rate of change of distance.
Average speed: The ratio of the total distance covered during a journey to the total time taken is called average speed.
Mathematically,$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.
The $SI$ unit of speed is $m/s$ $(ms^{-1})$,and it is a scalar quantity. Therefore,its value is always positive.

(N/A) Velocity: The rate of change of displacement of an object with respect to time is called velocity. It is a vector quantity,meaning it has both magnitude and direction. The $SI$ unit of velocity is $m/s$.
Average velocity: The average velocity of an object is defined as the total displacement divided by the total time interval taken to cover that displacement. Mathematically,$v_{avg} = \frac{\Delta x}{\Delta t}$,where $\Delta x$ is the total displacement and $\Delta t$ is the total time interval.

(N/A) The speed of an object is defined as the total distance traveled divided by the total time taken,while the magnitude of velocity is defined as the magnitude of the displacement divided by the total time taken.
Since the distance traveled is always greater than or equal to the magnitude of the displacement $(Distance \ge |Displacement|)$,it follows that the speed is always greater than or equal to the magnitude of the velocity.
Mathematically,this is expressed as: $\text{Speed} \ge |\text{Velocity}|$.
Equality holds true only when the object moves in a straight line without changing its direction.

(B) Average speed is defined as the total distance traveled divided by the total time taken. Since distance is a scalar quantity and is always non-negative,average speed is always non-negative.
Average velocity is defined as the total displacement divided by the total time taken. Displacement is a vector quantity and can be positive,negative,or zero depending on the direction of motion and the final position relative to the initial position.
Therefore,average velocity can be positive,negative,or zero.

(B) No,we do not agree.
Average speed is defined as the total distance traveled divided by the total time taken.
It is not simply the arithmetic mean of different speeds.
For example,if an object travels a distance $d_1$ with speed $v_1$ and distance $d_2$ with speed $v_2$,the average speed is $\frac{d_1 + d_2}{t_1 + t_2} = \frac{d_1 + d_2}{\frac{d_1}{v_1} + \frac{d_2}{v_2}}$,which is the harmonic mean of speeds if distances are equal,not the arithmetic mean.

(B) No,it is not possible.
Velocity is defined as a vector quantity that includes both speed and direction.
If the velocity of an object is constant,both its magnitude (speed) and its direction must remain unchanged.
Therefore,if the velocity is constant,the speed cannot change.

(B) The statement is $False$.
Speed is defined as the magnitude of velocity $(v = |\vec{v}|)$.
If the speed is zero, it implies that the magnitude of the velocity vector is zero, which means the velocity must be zero $(\vec{v} = 0)$.
Therefore, it is impossible for a particle to have zero speed and non-zero velocity simultaneously.