Q.Four position-time ($x$ versus $t$) graphs of a particle moving along a straight line are described below. In only ONE of them can the average velocity over the interval $(0, T)$ be made to vanish for a suitably chosen $T$. Which one is it?
Graph (a): at $t=0$ the position is negative (below the $t$-axis); the curve rises, crosses $x=0$ at some later time, reaches a positive maximum and then falls slightly, remaining positive.
Graph (b): at $t=0$ the position is positive; the curve rises to a maximum and then falls in an S-shape to a smaller positive value.
Graph (c): at $t=0$ the position is a large positive value; the curve decreases steadily (concave up) and levels off toward zero, staying positive.
Graph (d): the curve starts at the origin ($x=0$ at $t=0$), rises quickly and then flattens (concave down), approaching a constant positive value.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A motorist completes half revolution on a circular path of 60 m radius in one minute. His average speed (A) $1\ ms^{-1}$ (B) $2\ ms^{-1}$ (C) $3.14\ ms^{-1}$ (D) $4.14\ ms^{-1}$

   ›Reveal solutionSolution

   Average speed uses the actual path length travelled (the arc), not the straight-line displacement. Half a revolution covers $\pi r$, giving average speed $\pi\approx 3.14$ m/s.

   Concept and Intuition

   Average speed is distance travelled divided by time elapsed — it does not care about direction, unlike average velocity (which would use the displacement, i.e., the diameter here, and give a much smaller value). For half a revolution, the path length is half the circumference.

   Step-by-Step Solution

   1. Circumference of the circular path: $2\pi r = 2\pi(60) = 120\pi$ m.
   2. Half revolution ⇒ arc length travelled $= \dfrac{120\pi}{2} = 60\pi$ m $\approx 188.5$ m.
   3. Time taken $= 1$ minute $= 60$ s.
   4. Average speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{60\pi}{60} = \pi \approx 3.14\ \text{m/s}$.

   Common Mistakes

   • Using the diameter ($2r=120$ m, the displacement for a half revolution) instead of the arc length — that would compute average velocity magnitude ($120/60=2$ m/s), not average speed.
   • Forgetting to convert one minute to 60 seconds.

   ✓Final answer

   The correct option is (C) — $3.14\ ms^{-1}$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A Car travels first half of the distance with a velocity 'V' and second half of the distance with a velocity '3V', then the average velocity is (A) 2.0 V (B) 3.0 V (C) 4.0 V (D) 1.5 V

   ›Reveal solutionSolution

   Average velocity over two equal distances (not equal times) at speeds $V$ and $3V$ is the harmonic-mean-like combination $\dfrac{2v_1v_2}{v_1+v_2}=1.5V$.

   Concept and Intuition

   Average velocity is total displacement divided by total time — not the arithmetic mean of the two speeds, because the car spends more time at the slower speed. For equal distances $d$ at speeds $v_1,v_2$, the total distance is $2d$ and total time is $d/v_1+d/v_2$.

   Step-by-Step Solution

   1. Let total distance be $2d$; first half $d$ at speed $V$, second half $d$ at speed $3V$.
   2. Time for first half: $t_1=d/V$. Time for second half: $t_2=d/(3V)$.
   3. Total time: $t_1+t_2=\dfrac{d}{V}+\dfrac{d}{3V}=\dfrac{3d+d}{3V}=\dfrac{4d}{3V}$.
   4. Average velocity $=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{2d}{4d/(3V)}=\dfrac{2d\cdot3V}{4d}=\dfrac{6V}{4}=1.5V$.

   Common Mistakes

   • Taking the simple arithmetic mean $(V+3V)/2=2V$ — wrong because equal distances (not equal times) are covered at each speed.
   • Using the formula for equal-time averaging instead of equal-distance averaging.

   ✓Final answer

   The correct option is (D) — 1.5 V.

   ANSWER: D

3. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If a car travels 40% of the total distance with a speed $v_1$ and the remaining distance with a speed $v_2$, then average speed of the car is (A) $\dfrac{1}{2}\sqrt{v_1 v_2}$ (B) $\dfrac{v_1+v_2}{2}$ (C) $\dfrac{2v_1 v_2}{v_1+v_2}$ (D) $\dfrac{5v_1 v_2}{3v_1+2v_2}$

   ›Reveal solutionSolution

   This tests the formula for average speed as total distance over total time, not a simple average of speeds; the answer is (D).

   Concept and Intuition

   Average speed is never the arithmetic mean of two speeds unless the times (not distances) spent at each speed are equal. When distances (or fractions of distance) are given instead, you must add up the actual times taken for each segment.

   Step-by-Step Solution

   1. Let total distance be $D$. First segment: $0.4D$ at speed $v_1$, taking time $t_1=\dfrac{0.4D}{v_1}$.
   2. Second segment: remaining $0.6D$ at speed $v_2$, taking time $t_2=\dfrac{0.6D}{v_2}$.
   3. Average speed $=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{D}{t_1+t_2}=\dfrac{D}{\frac{0.4D}{v_1}+\frac{0.6D}{v_2}}=\dfrac{1}{\frac{0.4}{v_1}+\frac{0.6}{v_2}}$.
   4. Combine the fractions: $\dfrac{0.4}{v_1}+\dfrac{0.6}{v_2}=\dfrac{0.4v_2+0.6v_1}{v_1v_2}$.
   5. So average speed $=\dfrac{v_1v_2}{0.6v_1+0.4v_2}$. Multiplying numerator and denominator by 5 (to clear decimals): $\dfrac{5v_1v_2}{3v_1+2v_2}$.

   Common Mistakes

   • Averaging $v_1$ and $v_2$ directly (that's only valid for equal times, not equal distance fractions).
   • Arithmetic slip converting 0.4/0.6 into the 2/3 fraction form.

   ✓Final answer

   The correct option is (D) — $\dfrac{5v_1v_2}{3v_1+2v_2}$.

   ANSWER: D

4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A car is moving with a velocity of $4\ \mathrm{m\,s^{-1}}$ towards east. After a time of 4 s, if it is heading north-east with a velocity of $4\sqrt{2}\ \mathrm{m\,s^{-1}}$, then the average velocity of the car is (A) $2\sqrt{5}\ \mathrm{m\,s^{-1}}$ (B) $3\sqrt{5}\ \mathrm{m\,s^{-1}}$ (C) $4\sqrt{3}\ \mathrm{m\,s^{-1}}$ (D) $5\sqrt{3}\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   Under constant acceleration, average velocity is the vector average of the initial and final velocities, $(\vec v_1+\vec v_2)/2$. Computing this vector sum gives (A) $2\sqrt5$ m/s.

   Concept and Intuition

   For a body under constant acceleration $\vec a$, displacement is $\vec s=\vec v_1 t+\tfrac12\vec a t^2$, and average velocity is $\vec v_{avg}=\vec s/t=\vec v_1+\tfrac12\vec a t$. Since $\vec v_2=\vec v_1+\vec a t$, we can write $\tfrac12\vec a t=\tfrac12(\vec v_2-\vec v_1)$, so $\vec v_{avg}=\vec v_1+\tfrac12(\vec v_2-\vec v_1)=\dfrac{\vec v_1+\vec v_2}{2}$. This vector identity holds even though the direction of velocity changes, as long as the acceleration is constant.

   Step-by-Step Solution

   1. Set up axes: east $=\hat i$, north $=\hat j$.
   2. Initial velocity: $\vec v_1=4\hat i\ \mathrm{m\,s^{-1}}$ (due east).
   3. Final velocity: north-east means 45° between east and north, so $\vec v_2=4\sqrt2(\cos45^\circ\hat i+\sin45^\circ\hat j)=4\sqrt2\left(\tfrac{\sqrt2}{2}\hat i+\tfrac{\sqrt2}{2}\hat j\right)=4\hat i+4\hat j$.
   4. Average velocity $=\dfrac{\vec v_1+\vec v_2}{2}=\dfrac{(4+4)\hat i+(0+4)\hat j}{2}=4\hat i+2\hat j$.
   5. Magnitude: $|\vec v_{avg}|=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt5\ \mathrm{m\,s^{-1}}$.

   Common Mistakes

   • Simply subtracting speeds ($4\sqrt2-4$) instead of treating velocity as a vector.
   • Forgetting the 45° angle between 'east' and 'north-east' when resolving $\vec v_2$ into components.
   • Computing $|\vec v_2-\vec v_1|/t$ (average acceleration) instead of $(\vec v_1+\vec v_2)/2$ (average velocity) — these are different physical quantities.

   ✓Final answer

   The correct option is (A) — $2\sqrt{5}\ \mathrm{m\,s^{-1}}$.

   ANSWER: A

5. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.A body moving along a straight line path travels first 10 m distance in a time of 3 seconds and the next 10 m distance with a velocity of $5\ \mathrm{m\,s^{-1}}$. The average velocity of the body is (A) 4 kmph (B) 10.6 kmph (C) 18.2 kmph (D) 14.4 kmph

   ›Reveal solutionSolution

   Total distance 20 m over total time 5 s gives average velocity 4 m/s, which converts to 14.4 km/h. Answer: (D).

   Concept and Intuition

   Average velocity over a whole journey is always total displacement divided by total time, not simply an average of the different speeds used in each part. Whenever a journey is split into segments each with their own speed or time, the correct approach is to work out the time (or distance) for each segment separately, sum them, and only then divide.

   Step-by-Step Solution

   1. First segment: distance $= 10\ \text{m}$, time $= 3\ \text{s}$ (given directly).
   2. Second segment: distance $= 10\ \text{m}$, travelled at $5\ \mathrm{m\,s^{-1}}$, so time $= \dfrac{10}{5} = 2\ \text{s}$.
   3. Total distance $= 10 + 10 = 20\ \text{m}$.
   4. Total time $= 3 + 2 = 5\ \text{s}$.
   5. Average velocity $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{20}{5} = 4\ \mathrm{m\,s^{-1}}$.
   6. Convert to km/h: $4\ \mathrm{m\,s^{-1}} \times 3.6 = 14.4$ km/h.

   Common Mistakes

   • Simply averaging the two segment speeds arithmetically (which is only valid for equal times, not equal distances covered at different speeds).
   • Forgetting to convert the final answer from m/s to km/h (which is why 4 kmph appears as a distractor option, from skipping the $\times 3.6$ conversion).

   ✓Final answer

   The correct option is (D) — 14.4 kmph.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.A particle moving along a straight line covers the first half of the distance with a speed of $3\ \mathrm{m\,s^{-1}}$, the other half of the distance is covered in two equal time intervals with speeds of $4.5\ \mathrm{m\,s^{-1}}$ and $7.5\ \mathrm{m\,s^{-1}}$ respectively, then the average speed of particle during the motion is (A) $4.0\ \mathrm{m\,s^{-1}}$ (B) $5.0\ \mathrm{m\,s^{-1}}$ (C) $5.5\ \mathrm{m\,s^{-1}}$ (D) $4.8\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   This tests computing average speed as total distance divided by total time, carefully handling a first half defined by distance and a second half defined by equal time intervals. The average speed is $4.0\text{ m/s}$, option (A).

   Concept and Intuition

   Average speed is always $\dfrac{\text{total distance}}{\text{total time}}$ — never a simple average of the individual speeds unless the time intervals happen to be equal for all segments. Here the first half of the distance takes one particular time, while the second half of the distance is split into two equal time intervals at different speeds; we must find the time taken for each part separately using $t=\dfrac{\text{distance}}{\text{speed}}$ or the given time-split information, then add them up.

   Step-by-Step Solution

   1. Let the total distance be $2d$, so each half is $d$.
   2. First half: distance $d$ at speed $3\text{ m/s}$, so time $t_1=\dfrac{d}{3}$.
   3. Second half: distance $d$ covered in two equal time intervals $\tau$ each, at speeds $4.5\text{ m/s}$ and $7.5\text{ m/s}$. So distance covered $=4.5\tau+7.5\tau=12\tau=d \Rightarrow \tau=\dfrac{d}{12}$.
   4. Total time for second half $=2\tau=\dfrac{d}{6}$.
   5. Total time $=t_1+2\tau=\dfrac{d}{3}+\dfrac{d}{6}=\dfrac{2d+d}{6}=\dfrac{d}{2}$.
   6. Average speed $=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{2d}{d/2}=4\text{ m/s}$.

   Common Mistakes

   • Averaging the three speeds $(3+4.5+7.5)/3=5\text{ m/s}$ directly — this ignores that the time spent at each speed isn't equal, giving the wrong (but tempting) distractor (B).
   • Assuming the two equal time intervals in the second half also correspond to equal distances (they don't — the distances differ since the speeds differ).

   ✓Final answer

   The correct option is (A) — $4.0\ \mathrm{m\,s^{-1}}$.

   ANSWER: A

7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The displacement

   (x) and time (t) graph of a particle moving along a straight line is shown in the figure. The average velocity of the particle in the time of 10 s is [FIGURE] (a graph of x(m) vs t(s): the line starts at x = 80 m at t = 0, decreases linearly to x = 20 m at t = 6 s, then increases linearly to x = 60 m at t = 10 s) (A) $2\ \mathrm{m\,s^{-1}}$ (B) $4\ \mathrm{m\,s^{-1}}$ (C) $6\ \mathrm{m\,s^{-1}}$ (D) $8\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   Average velocity is net displacement over total time — the intermediate dip in the graph doesn't matter. Answer: 2 m/s.

   Concept and Intuition

   A common trap is to try to add up distances covered on each segment of the V-shaped graph. But average velocity (as opposed to average speed) only cares about the straight-line displacement between the initial and final positions and the total elapsed time.

   Step-by-Step Solution

   1. From the graph: at $t=0$, $x=80$ m; at $t=10$ s, $x=60$ m.
   2. Net displacement $= x(10) - x(0) = 60 - 80 = -20$ m.
   3. Average velocity $= \dfrac{\Delta x}{\Delta t} = \dfrac{-20}{10} = -2\ \mathrm{m\,s^{-1}}$.
   4. The magnitude of average velocity is $2\ \mathrm{m\,s^{-1}}$.

   Common Mistakes

   • Confusing average velocity with average speed (which would require adding the distance decreasing from 80→20 and increasing from 20→60, giving a different, larger number).
   • Using only one segment's slope instead of the overall displacement.

   ✓Final answer

   The correct option is (A) — $2\ \mathrm{m\,s^{-1}}$.

   ANSWER: A

8. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If a person moving along a straight line path covers first half distance with velocity 'V$_1$' and the next half distance with velocity 'V$_2$', then the average velocity of the person is (A) $\dfrac{V_1+V_2}{2}$ (B) $\dfrac{V_1+V_2}{2\sqrt{V_1V_2}}$ (C) $\dfrac{2}{\dfrac{1}{V_1}+\dfrac{1}{V_2}}$ (D) $\dfrac{V_1V_2}{V_1+V_2}$

   ›Reveal solutionSolution

   Equal distances covered at different speeds average to the harmonic mean of the speeds, not the arithmetic mean — here that harmonic mean simplifies to option (D).

   Concept and Intuition

   Averaging speeds is only a simple arithmetic mean when the times spent at each speed are equal. When instead the distances are equal (as here — 'first half distance', 'next half distance'), the body spends more time at the slower speed, which pulls the average down toward the smaller value. This is exactly the situation that produces a harmonic mean rather than an arithmetic mean.

   Step-by-Step Solution

   1. Let the total distance be $2s$, so each half is $s$.
   2. Time for the first half: $t_1=\dfrac{s}{V_1}$.
   3. Time for the second half: $t_2=\dfrac{s}{V_2}$.
   4. Average velocity $=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{2s}{\frac{s}{V_1}+\frac{s}{V_2}}=\dfrac{2}{\frac1{V_1}+\frac1{V_2}}=\dfrac{2V_1V_2}{V_1+V_2}$.
   5. Comparing with the printed options, option (C) $\dfrac{2}{\frac1{V_1}+\frac1{V_2}}$ is the same harmonic-mean expression written in a different (unreduced) form, while (D) $\dfrac{V_1V_2}{V_1+V_2}$ is exactly half of the true value — so on strict algebra option (C) is the exact match.

   Common Mistakes

   • Averaging $V_1$ and $V_2$ arithmetically (option A) — valid only for equal time intervals, not equal distances.
   • Dropping the factor of 2 when simplifying the harmonic-mean formula.

   ✓Final answer

   The correct option is (C) — $\dfrac{2}{\dfrac{1}{V_1}+\dfrac{1}{V_2}}$, the harmonic mean of $V_1$ and $V_2$.

   ANSWER: C

9. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
   Q.A person is running with an uniform velocity towards a flyover. He takes 5 s to reach the flyover from a reference point and takes 50 s to cross the flyover from the same reference point. If the length of the flyover is 1000 m then his velocity is nearly (A) 83.1 kmph (B) 80.0 kmph (C) 75.4 kmph (D) 85.2 kmph

   ›Reveal solutionSolution

   Both given times are measured from the same reference point, so the time spent actually crossing the flyover is the difference of the two times, not either one directly.

   Concept and Intuition

   The person moves at constant (uniform) velocity throughout. "Reaching the flyover" and "crossing the flyover" are both timed from the same starting reference point, so:

   • $t_1 = 5$ s: time to reach the start of the flyover.
   • $t_2 = 50$ s: time to reach the end of the flyover (i.e., to have crossed it), from the same reference point.

   The actual time spent traversing the flyover's length is $t_2 - t_1$.

   Step-by-Step Solution

   1. Time to cross the flyover $= t_2 - t_1 = 50 - 5 = 45$ s.
   2. Length of flyover $= 1000$ m.
   3. Velocity $v = \dfrac{1000\text{ m}}{45\text{ s}} = 22.22$ m/s.
   4. Convert to km/h: $22.22 \times 3.6 = 80.0$ kmph.

   Common Mistakes

   • Using $t=50$ s directly with 1000 m (forgetting both times share the same reference point), which gives the wrong value 20 m/s.
   • Forgetting to subtract the initial 5 s before dividing.

   ✓Final answer

   The correct option is (B) — 80.0 kmph.

   ANSWER: B

10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.A car covers a distance at speed of $60\ \mathrm{km\,h^{-1}}$. It returns and comes back to the original point moving at a speed of V. If the average speed for the round trip is $48\ \mathrm{km\,h^{-1}}$, then the magnitude of V is (A) $40\ \mathrm{km\,h^{-1}}$ (B) $36\ \mathrm{km\,h^{-1}}$ (C) $44\ \mathrm{km\,h^{-1}}$ (D) $32\ \mathrm{km\,h^{-1}}$

    ›Reveal solutionSolution

    For equal-distance round trips at two speeds, average speed is the harmonic mean, not the arithmetic mean; solving gives $V=40\ \mathrm{km\,h^{-1}}$.

    Concept and Intuition

    Average speed = total distance / total time. For a trip covering the same distance $d$ each way at speeds $v_1$ and $v_2$, total time $=\dfrac{d}{v_1}+\dfrac{d}{v_2}$, and total distance $=2d$, giving $\bar v = \dfrac{2v_1v_2}{v_1+v_2}$ — the harmonic mean, always less than the arithmetic mean.

    Step-by-Step Solution

    1. Let distance one-way be $d$. Time going $=d/60$, time returning $=d/V$.
    2. Average speed $=\dfrac{2d}{d/60+d/V}=\dfrac{2}{\frac1{60}+\frac1V}=\dfrac{2\cdot60V}{60+V}$.
    3. Set this equal to 48: $\dfrac{120V}{60+V}=48$.
    4. $120V = 48(60+V) = 2880+48V$.
    5. $72V=2880 \Rightarrow V=40\ \mathrm{km\,h^{-1}}$.

    Common Mistakes

    • Using the arithmetic mean $(v_1+v_2)/2=48$ directly (giving $V=36$, a distractor option) instead of the correct harmonic-mean relation, since the two legs take different times over the same distance.

    ✓Final answer

    The correct option is (A) — $40\ \mathrm{km\,h^{-1}}$.

    ANSWER: A

11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.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]. It's average speed during this motion is [FIGURE] (x-t graph: (0,1) to (1,2) to (2,3) to (3,3) to (4,2) to (5,3), straight line segments) (A) $0.4\ \mathrm{ms^{-1}}$ (B) $1.0\ \mathrm{ms^{-1}}$ (C) $0.8\ \mathrm{ms^{-1}}$ (D) $0.6\ \mathrm{ms^{-1}}$

    ›Reveal solutionSolution

    Average speed uses total distance (always positive, path length), not net

    displacement — read the distance off each segment of the x-t graph and divide

    by the total time.

    Concept and Intuition

    On an x-t graph, the slope of each segment is the velocity during that

    interval, but average speed cares only about how much ground was actually

    covered — direction reversals still add positively to the distance. So we must

    look at each of the five 1-second segments individually and add up

    $|\Delta x|$ for each, rather than just taking $|x(5)-x(0)|$.

    Step-by-Step Solution

    1. Read the vertices from the graph: $(0,1)\to(1,2)\to(2,3)\to(3,3)\to(4,2)\to(5,3)$.
    2. Segment 1 ($t=0\to1$): $x:1\to2$, distance $=1$ m (motion in $+x$).
    3. Segment 2 ($t=1\to2$): $x:2\to3$, distance $=1$ m (motion in $+x$).
    4. Segment 3 ($t=2\to3$): $x:3\to3$, distance $=0$ (particle momentarily at rest).
    5. Segment 4 ($t=3\to4$): $x:3\to2$, distance $=1$ m (motion reverses, in $-x$).
    6. Segment 5 ($t=4\to5$): $x:2\to3$, distance $=1$ m (motion in $+x$ again).
    7. Total distance $= 1+1+0+1+1 = 4$ m. Total time $=5$ s.
    8. Average speed $=\dfrac{4\ \text{m}}{5\ \text{s}} = 0.8\ \mathrm{ms^{-1}}$.

    Common Mistakes

    • Computing average velocity instead: net displacement is $x(5)-x(0)=3-1=2$ m, giving $0.4\ \mathrm{ms^{-1}}$ (option A) — that's the trap, not what's asked.
    • Missing the zero-distance flat segment ($t=2\to3$) or the direction reversal ($t=3\to4$), which silently cancels in a displacement calculation but must still be counted in distance.

    ✓Final answer

    The correct option is (C) — $0.8\ \mathrm{ms^{-1}}$.

    ANSWER: C

12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.A biker travels $\frac{1}{3}$ of the distance L with speed $v_1$ and $\frac{2}{3}$ of the distance with speed $v_2$. Then the average speed is (A) $\dfrac{v_1 v_2}{v_1 + v_2}$ (B) $\dfrac{3 v_1 v_2}{2 v_1 + v_2}$ (C) $\dfrac{3 v_1 v_2}{v_1 + 2 v_2}$ (D) $\dfrac{v_1 + v_2}{v_1 v_2}$

    ›Reveal solutionSolution

    Average speed = total distance / total time; working through the two legs gives $\frac{3v_1v_2}{2v_1+v_2}$.

    Concept and Intuition

    Average speed over a trip with different speeds on different legs is NOT the simple average of the speeds — it must be computed as total distance divided by total time, weighting by how long each leg actually takes.

    Step-by-Step Solution

    1. First leg: distance $=L/3$, speed $=v_1$, so time $t_1=\dfrac{L}{3v_1}$.
    2. Second leg: distance $=2L/3$, speed $=v_2$, so time $t_2=\dfrac{2L}{3v_2}$.
    3. Total time $=t_1+t_2=\dfrac{L}{3v_1}+\dfrac{2L}{3v_2}=\dfrac{L}{3}\left(\dfrac{1}{v_1}+\dfrac{2}{v_2}\right)=\dfrac{L(v_2+2v_1)}{3v_1v_2}$.
    4. Average speed $=\dfrac{L}{t_1+t_2}=\dfrac{L\cdot3v_1v_2}{L(v_2+2v_1)}=\dfrac{3v_1v_2}{2v_1+v_2}$.

    Common Mistakes

    • Using the arithmetic mean $(v_1+v_2)/2$ or harmonic mean for equal-time (not equal-distance) legs — the weighting here is by distance fraction, not time.
    • Mixing up which speed goes with which distance fraction.

    ✓Final answer

    The correct option is (B) — $\dfrac{3v_1v_2}{2v_1+v_2}$.

    ANSWER: B