The trap first: aspirants routinely treat average velocity and average speed as interchangeable. They are not. For a round trip — same path, same total distance — average speed is positive; average velocity is zero. That single distinction costs marks when a distractor offers the arithmetic mean of two speeds as the "average velocity."
Speed is a scalar: the rate at which distance (path length) increases. Velocity is a vector: the rate at which displacement changes. NCERT Class 11 Physics Chapter 2, page 14 defines speed as the magnitude of velocity only for the instantaneous case. For averages, the two diverge whenever the path is not straight or involves a direction change.
Definitions to lock in:
Where NEET tests this: questions give a multi-leg journey (e.g., half the distance at speed v₁, half at v₂) and ask for average speed. The trap distractor is (v₁ + v₂)/2 — the arithmetic mean — which is wrong. The correct formula is the harmonic mean: 2v₁v₂/(v₁ + v₂). For equal-time legs, the arithmetic mean is correct; for equal-distance legs, the harmonic mean is correct. Mixing these up is the core confusion.
Watch-out: if a question says "find the average velocity" for a closed-path journey, the answer is zero regardless of speed variations along the way. Displacement is zero; velocity = displacement/time = 0.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
A car travels from A to B (distance 120 km) at 60 km/h, then returns from B to A at 40 km/h. What is the average speed for the entire trip?
Answer: B. Total distance = 240 km. Time for A→B = 120/60 = 2 h; time for B→A = 120/40 = 3 h. Total time = 5 h. Average speed = 240/5 = 48 km/h. This is the harmonic mean 2(60)(40)/(60+40) = 4800/100 = 48 km/h (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because 50 km/h is the arithmetic mean (60+40)/2 = 50, which applies only when equal TIME is spent at each speed, not equal distance (trap: confusing equal-distance with equal-time averaging).
Why C is wrong: C is wrong because 0 is the average VELOCITY (displacement = 0 for a round trip), not the average speed. Average speed uses total path length, which is 240 km, not zero (trap: conflating average velocity with average speed).
Why D is wrong: D is wrong because 100 km/h is simply the sum of the two speeds. Average speed is never the sum of individual speeds — it is total distance divided by total time.
A runner completes one full lap of a circular track of circumference 400 m in 80 s. What are the average speed and average velocity for the complete lap?
Answer: B. Average speed = total path length / total time = 400/80 = 5 m/s. For a complete lap, displacement = 0 (start = finish), so average velocity = 0/80 = 0 (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because while average speed is correctly 5 m/s, average velocity cannot also be 5 m/s — the runner returns to the starting point, so displacement is zero and average velocity is 0 (trap: treating speed and velocity as identical).
Why C is wrong: C is wrong because average speed is not zero — the runner covered 400 m of path length. Only displacement (and hence average velocity) is zero for a closed loop.
Why D is wrong: D is wrong because it reverses the two quantities. Average velocity is 0 (zero displacement), and average speed is 5 m/s (400 m covered). This option swaps them.
Which of the following statements is correct?
Answer: C. At any instant, speed is defined as the magnitude of velocity. This holds by definition (NCERT Class 11 Physics Chapter 2, page 14). However, this equality does NOT extend to their averages over a finite interval.
Why A is wrong: A is wrong because speed is a scalar quantity defined as path length over time — both are non-negative, so average speed is always ≥ 0. Only velocity (a vector) can be negative.
Why B is wrong: B is wrong because average velocity equals average speed only when the motion is along a straight line in one direction (no reversal). For any path with direction change, average speed > |average velocity| (trap: assuming the two are always equal).
Why D is wrong: D is wrong because for any non-straight path, total path length > |total displacement|, so average speed > |average velocity|. Equality holds only for unidirectional straight-line motion.
A body travels half the total distance at speed v₁ and the remaining half at speed v₂. The average speed for the whole journey is:
Answer: B. Let total distance = 2d. Time for first half = d/v₁; time for second half = d/v₂. Average speed = 2d / (d/v₁ + d/v₂) = 2v₁v₂/(v₁ + v₂), the harmonic mean (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because the arithmetic mean (v₁+v₂)/2 gives the correct average speed only when equal TIME is spent at each speed, not equal distance. For equal-distance legs, the harmonic mean applies (trap: confusing equal-distance and equal-time conditions).
Why C is wrong: C is wrong because the geometric mean √(v₁v₂) does not arise in any standard averaging of speeds over distance or time segments. It has no physical basis here.
Why D is wrong: D is wrong because (v₁²+v₂²)/(v₁+v₂) does not correspond to any standard speed-averaging formula. It is dimensionally correct but algebraically wrong — it overestimates the true average.
A body travels half the total time at speed v₁ and the remaining half time at speed v₂. The average speed is:
Answer: B. Let total time = 2t. Distance in first half = v₁t; in second half = v₂t. Average speed = (v₁t + v₂t) / 2t = (v₁ + v₂)/2, the arithmetic mean (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because the harmonic mean 2v₁v₂/(v₁+v₂) is the correct average speed when equal DISTANCES are covered at each speed, not equal times (trap: applying the equal-distance formula to an equal-time problem).
Why C is wrong: C is wrong because the geometric mean has no standard role in speed averaging over time or distance. It is a mathematical mean but not the physically correct one here.
Why D is wrong: D is wrong because v₁v₂/(v₁+v₂) is half the harmonic mean — neither the harmonic mean itself nor the correct arithmetic mean.
A car moves along a straight road: 30 km at 30 km/h, then 30 km at 60 km/h. The average speed is:
Answer: B. Total distance = 60 km. Time₁ = 30/30 = 1 h; Time₂ = 30/60 = 0.5 h. Total time = 1.5 h. Average speed = 60/1.5 = 40 km/h. Alternatively: harmonic mean = 2(30)(60)/(30+60) = 3600/90 = 40 km/h (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because 45 km/h is the arithmetic mean (30+60)/2 = 45. This would be correct only if equal time were spent at each speed. Here equal distances were covered, so the harmonic mean applies (trap: using arithmetic mean for equal-distance legs).
Why C is wrong: C is wrong. 48 km/h would be the harmonic mean of 60 and 40, not of 30 and 60. This comes from misreading the given speeds.
Why D is wrong: D is wrong. 36 km/h does not match any standard averaging of 30 and 60 km/h. It may arise from an arithmetic error such as dividing 60 km by 5/3 h incorrectly.
A particle moves along the x-axis: from x = 0 to x = 5 m in 2 s, then from x = 5 m to x = 2 m in 1 s. The average velocity over the entire 3 s interval is:
Answer: B. Total displacement = final position − initial position = 2 − 0 = 2 m. Total time = 3 s. Average velocity = 2/3 m/s (NCERT Class 11 Physics Chapter 2, page 14). Note: average speed would be (5 + 3)/3 = 8/3 m/s — a different quantity.
Why A is wrong: A is wrong because 8/3 m/s is the average SPEED (total path length 8 m divided by 3 s), not the average velocity. Average velocity uses displacement (2 m), not path length (trap: confusing path length with displacement).
Why C is wrong: C is wrong because 7/3 m/s does not correspond to either displacement/time or path-length/time for this problem. It likely arises from miscalculating the return leg distance as 2 m instead of 3 m.
Why D is wrong: D is wrong because 1 m/s would require a displacement of 3 m over 3 s. The actual displacement is 2 m (from x = 0 to x = 2), giving 2/3 m/s.
Two straight-line position-time graphs for particles P and Q make angles of 30° and 60° respectively with the time axis. The ratio of velocities v_P : v_Q is:
Answer: C. Velocity = slope of x-t graph = tan(angle with time axis). v_P = tan 30° = 1/√3; v_Q = tan 60° = √3. Ratio = (1/√3)/√3 = 1/3, so v_P : v_Q = 1 : 3 (anchored to PYQ pattern from 2022 session; NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because 1:√3 would be the ratio of sin 30° to sin 60° (= 0.5 : 0.866). But velocity is the SLOPE of the x-t graph, which requires tan, not sin (trap from pattern NEET pattern: position time graph slope: using sin or cos instead of tan).
Why B is wrong: B is wrong because √3:1 reverses the correct ratio. tan 30° < tan 60°, so v_P < v_Q, meaning P's velocity is the smaller one, not the larger.
Why D is wrong: D is wrong because 1:1 would mean both particles have the same velocity, which requires equal slopes — but 30° and 60° give different tan values.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Given
• Segment 1: angle with time axis = 45° (exact), duration = 4 s, starts at x = 0.• Segment 2: angle with time axis = 30° (exact) below the horizontal (falling), duration = 6 s.
Required
(a) Velocity in each segment. (b) Average speed over 10 s. (c) Average velocity over 10 s.
Concept
Velocity = slope of x-t graph = tan(angle with time axis). For a falling segment, the slope is negative. Average speed = total path length / total time. Average velocity = net displacement / total time.
Formula
v = tan(θ) for straight-line x-t graph.
Average speed = (d₁ + d₂) / (t₁ + t₂).
Average velocity = (x_final − x_initial) / (t₁ + t₂).
Substitution
Segment 1: v₁ = tan 45° = 1 m/s. Distance covered = 1 × 4 = 4 m. Position at t = 4 s: x = 4 m.
Segment 2: slope = −tan 30° = −1/√3 m/s ≈ −0.577 m/s (negative: moving back). Distance covered = (1/√3) × 6 = 6/√3 = 2√3 m ≈ 3.464 m. Position at t = 10 s: x = 4 − 2√3 m.
Calculation
(a) v₁ = +1 m/s; v₂ = −1/√3 m/s ≈ −0.577 m/s.
(b) Total path length = 4 + 2√3 = 4 + 3.464 = 7.464 m. Average speed = 7.464/10 ≈ 0.746 m/s.
(c) Net displacement = 4 − 2√3 = 4 − 3.464 = 0.536 m. Average velocity = 0.536/10 ≈ 0.054 m/s.
Note on exact constants: The angles 45° and 30° are exact geometric values; the integers 4 s and 6 s are exact counting durations. These do not limit significant figures. The irrational value √3 = 1.732… is exact. Reported answers are rounded for clarity.
Final answer
(a) Segment 1 velocity: 1.00 m/s; Segment 2 velocity: −0.577 m/s.
(b) Average speed ≈ 0.746 m/s.
(c) Average velocity ≈ 0.054 m/s.
Note the large gap between average speed and average velocity — the particle nearly returns to its starting point, making displacement small while total path length remains substantial.
Common trap
Confusing average speed with average velocity. A student who reports 0.054 m/s as the "average speed" has used displacement instead of path length. Conversely, reporting 0.746 m/s as "average velocity" uses path length instead of displacement. The two are equal only for unidirectional straight-line motion.
Also: using sin 45° or sin 30° instead of tan to extract velocity from the graph slope (the trig-function trap from the position-time-graph pattern).
Similar NEET-style question
A particle's x-t graph is a straight line making 60° with the time axis for the first 5 s, then a horizontal line for the next 5 s. Find the average speed and average velocity over the full 10 s.
*(Answer: velocity in first segment = tan 60° = √3 m/s; distance = 5√3 m. Second segment: stationary. Average speed = 5√3/10 = √3/2 m/s ≈ 0.866 m/s. Average velocity = 5√3/10 = √3/2 m/s ≈ 0.866 m/s. Here they are equal because the motion is unidirectional.)*
---
The instantaneous velocity v at an instant t is the limit of the average velocity Δx/Δt as Δt → 0: v = dx/dt. Instantaneous speed is the magnitude of instantaneous velocity. (Note: in the new edition, average velocity and average speed appear within section 2.2 rather than as a separate section.)
-- NCERT Class 11 Physics, Ch. 2, p. 14These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Average velocity = total displacement / total time (vector). Average speed = total path length / total time (scalar, always >= |average velocity|). For round-trip motion, average velocity is zero; average speed is not.
Distractor offers (v1+v2)/2 instead of total-distance/total-time.
More in Kinematics: 13 exam traps and mistakes · 6 formulas · 9 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
uses sin or cos instead of tan
Trig confusion under time pressure
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →