Two different questions, one clock. “How far did you end up from where you started?” gives
average velocity. “What does the odometer say?” gives average speed.
NCERT opens the chapter with this distinction and then never lets it go, because almost every
error later in kinematics traces back to it. Displacement is a vector: it has a sign and it only cares about
two points, the start and the finish. Path length is a scalar: it never decreases, and it counts every metre
travelled, including the ones you retraced.
animated Walk 5 m out and 5 m back. The odometer climbs to 10 m and stays there. The displacement climbs to 5 m and then falls back to zero. Same journey, same 10 seconds, two entirely different averages.
The inequality that never breaks
The shortest route between two points is the straight line between them, so the path length can never be
less than the magnitude of the displacement. Divide both by the same time and you get:
average speed ≥ |average velocity|
Equality holds in exactly one case — when the body never turns around. That is a useful test in
reverse: if a question tells you the two are equal, it has just told you the motion was one-way.
The asymmetry NEET keeps harvesting
Over an interval the two quantities can differ, and the inequality points one way. At an
instant they are always equal: instantaneous speed = |instantaneous velocity|, with no inequality at
all. In a vanishingly small interval there is simply no time to double back, so the path travelled and the
displacement become the same thing.
NCERT Example 2.1 — the source question
Everything in this topic is a variation on the example NCERT sets first: a car runs out and comes partly
back, and the two averages come apart the moment it turns.
animated The car reaches P (360 m) in 18 s, then reverses to Q (240 m) in 6 s. Over O→P nothing has reversed, so both averages are 20 m/s. Over the whole trip the displacement is only 240 m while the odometer reads 480 m — and the averages split into 10 and 20 m/s.
Reading it off a velocity–time graph
On a v–t graph, area below the time axis counts as negative displacement. Displacement is the
signed area; distance is the sum of the areas with all signs made positive. So the recipe is always:
find where v = 0, split the interval there, and only then integrate.
animated v = 6 − 2t. The particle runs forward until t = 3 s, then reverses. Signed areas: +9 and −4, giving a displacement of 5 m. Unsigned: 9 and 4, giving a distance of 13 m. Two averages, 1 and 2.6 m/s.
Equal distances or equal times?
The single most common calculation in this topic is the two-speed journey, and the answer depends
entirely on which quantity is split in half:
The journey is split into…
Average speed
Name
two equal distances
2v₁v₂/(v₁ + v₂)
harmonic mean
two equal times
(v₁ + v₂)/2
arithmetic mean
Equal distances always give the smaller answer, because you spend longer travelling at the slower speed
and it therefore carries more weight. For 40 and 60 km h⁻¹ the two answers are 48 and 50 —
close enough that both will be in the options, far enough apart to lose the mark.
2 · Formula sheet
Average velocity
v̄ = Δx/Δt
Signed. Uses only the endpoints — the path in between is irrelevant.
Average speed
total path length / Δt
Never negative. The odometer reading divided by the clock.
The inequality
average speed ≥ |v̄|
Equality only when the motion never reverses.
At an instant
instantaneous speed = |v|
Always equal — there is no time to double back in a vanishing interval.
Equal distances, two speeds
v̄ = 2v₁v₂/(v₁ + v₂)
Harmonic mean. Always less than the arithmetic mean.
Equal distances, n speeds
v̄ = n / Σ(1/vi)
The general harmonic mean.
Equal times, two speeds
v̄ = (v₁ + v₂)/2
Arithmetic mean.
Equal times, n speeds
v̄ = (Σvi)/n
Ordinary average.
Uniform acceleration only
v̄ = (u + v)/2
A special case, not a definition. Equals the mid-time velocity.
Velocity at the mid-point
√[(u² + v²)/2]
Root-mean-square. Always larger than the mid-time value.
Ratio of the two averages
path length / |displacement|
The time cancels — never compute either average for a ratio question.
Complete up-and-down flight
v̄ = 0 ; average speed = u/2
The speed falls linearly from u to 0 and back.
Closed path of any shape
v̄ = 0
One full lap of a circle, a square, anything.
From a graph
v̄ = slope of the chord
Instantaneous velocity is the slope of the tangent. Chord vs tangent is the whole distinction.
From a v–t graph
Δx = signed area ; distance = Σ|areas|
Split the integral wherever v changes sign.
3 · Exceptions and traps
Trap — average speed is not the average of the speeds
Adding speeds and dividing by how many there are is correct only when each speed was held for the
same length of time. Whenever the question splits the distance, you must go back to
total distance ÷ total time.
Trap — ‘half the journey’
English is ambiguous here; physics is not. Half the distance and half the time give different answers.
Find the word — distance, path, way, journey, time — before choosing a formula.
Trap — (u+v)/2 used where acceleration is not uniform
That formula is the average velocity only for constant acceleration, because only then does the
velocity change linearly with time. For any other motion, return to the definition
Δx/Δt.
Trap — integrating through a reversal
∫v dt over the whole interval gives displacement, never distance. If v
changes sign inside the interval, the negative part has already cancelled part of the positive part. Split
at every instant where v = 0.
Exception — zero average velocity does not mean rest
Any journey ending where it began has zero average velocity, however fast it was. To conclude that a
body was at rest you need the average speed to be zero.
Exception — the ratio question
When asked for average speed ÷ |average velocity|, the time cancels and the answer is simply
path length ÷ |displacement|. Computing either average is wasted effort.
4 · Numbers to remember
Situation
Result
NCERT Example 2.1, O→P
both averages 20 m s⁻¹
NCERT Example 2.1, O→P→Q
10 m s⁻¹ and 20 m s⁻¹
Equal distances at 40 and 60 km h⁻¹
48 km h⁻¹
Equal times at 40 and 60 km h⁻¹
50 km h⁻¹
Equal distances at 20, 30, 60 km h⁻¹
30 km h⁻¹
One full circular lap
v̄ = 0, average speed = 2πR/T
Half a circular lap
ratio of averages = π/2 ≈ 1.57
Complete vertical throw at speed u
v̄ = 0, average speed = u/2
Mid-time vs mid-point velocity
(u+v)/2 vs √[(u²+v²)/2]
5 · Scientists NEET names
Nicole Oresme (c. 1320–1382)
Four centuries before Newton, Oresme proved the mean speed theorem: a uniformly accelerating body
covers the same distance as one moving steadily at its average velocity, (u+v)/2.
He argued it from the area of a trapezium — the same picture NCERT uses in Fig. 2.5. Galileo later
used the result to derive free fall.
Galileo Galilei (1564–1642)
Named in the chapter for making the first quantitative studies of accelerated motion. The distinction
between distance and displacement is implicit in his work on the inclined plane, where he needed the path
along the slope and the drop in height to be kept separate.
6 · Sixteen worked questions
Built around NCERT Example 2.1 and the Summary point 1. No
previous-year attributions are claimed — items on this topic recur every year but under too many
different numberings to label confidently.
Q1NEET-style
A car moves along a straight line from O to P, a distance of 360 m, in 18 s, and then returns from P to Q in 6.0 s, where Q is 240 m from O. The average velocity and average speed of the car for the whole journey O → P → Q are
(a) 10 m s⁻¹ and 20 m s⁻¹
(b) 20 m s⁻¹ and 10 m s⁻¹
(c) 10 m s⁻¹ and 10 m s⁻¹
(d) 20 m s⁻¹ and 20 m s⁻¹
Given
OP = 360 m in 18 s; PQ = 120 m in 6 s; total time 24 s; final position 240 m from O
Asked
average velocity and average speed for O → P → Q
Concept to use
Average velocity uses the net displacement (start to finish, with sign). Average speed uses the total path length (odometer reading). Both are divided by the same total time.
Formula or rule
v̄ = Δx/Δt ; average speed = total path length/Δt
Baby steps
Displacement = 240 − 0 = 240 m.
Path length = 360 + 120 = 480 m.
Total time = 18 + 6 = 24 s.
v̄ = 240/24 = 10 m s⁻¹.
average speed = 480/24 = 20 m s⁻¹.
Answer
10 m s⁻¹ and 20 m s⁻¹
Shortcut
The two numbers differ by exactly the factor 480/240 = 2, because the ratio of the two answers is always path length ÷ |displacement|. Compute that ratio once and you get the second answer free.
Trap
Using 360 m for both. The car turned around, so the odometer and the ruler disagree from that moment on. This is NCERT Example 2.1 — the source question for this whole topic.
Q2NEET-style
A car covers the first half of a journey at 40 km h⁻¹ and the second half at 60 km h⁻¹. Its average speed for the whole journey is
(a) 24 km h⁻¹
(b) 50 km h⁻¹
(c) 48 km h⁻¹
(d) 100 km h⁻¹
Given
equal distances at 40 and 60 km h⁻¹
Asked
average speed
Concept to use
Average speed is total distance over total time, never the average of the speeds. Equal distances means you spend longer at the slower speed, so the slower speed carries more weight — the harmonic mean.
Formula or rule
v̄ = 2v₁v₂/(v₁ + v₂)
Baby steps
Let each half be d. Time = d/40 + d/60 = (3d + 2d)/120 = d/24.
Total distance = 2d.
v̄ = 2d ÷ (d/24) = 48 km h⁻¹.
Answer
48 km h⁻¹
Shortcut
Harmonic mean: 2(40)(60)/100 = 48. It is always below the arithmetic mean of 50, so if you see both numbers in the options the smaller one is right for equal distances.
Trap
Answering 50, the arithmetic mean. That is the answer to a different question — equal times at the two speeds. The words ‘half the journey’ are ambiguous in English but never in physics: check whether it is half the distance or half the time.
Q3NEET-style
A car travels for the first half of the time at 40 km h⁻¹ and for the second half of the time at 60 km h⁻¹. Its average speed is
(a) 24 km h⁻¹
(b) 48 km h⁻¹
(c) 120 km h⁻¹
(d) 50 km h⁻¹
Given
equal times at 40 and 60 km h⁻¹
Asked
average speed
Concept to use
With equal times, each speed contributes equally to the total distance, so the weighting is even and the answer is the ordinary arithmetic mean.
Formula or rule
v̄ = (v₁ + v₂)/2
Baby steps
Let each half be t. Distance = 40t + 60t = 100t.
Total time = 2t.
v̄ = 100t/2t = 50 km h⁻¹.
Answer
50 km h⁻¹
Shortcut
Equal times → arithmetic mean. Equal distances → harmonic mean. Two words, two formulas; this pair of questions is set together in almost every mock test.
Trap
Reaching for the harmonic mean out of habit. Compare this question with the previous one word by word — the only change is ‘journey’ to ‘time’, and the answer changes from 48 to 50.
Q4NEET-style
A body covers three equal successive distances with speeds 20, 30 and 60 km h⁻¹. Its average speed over the whole journey is
(a) 25 km h⁻¹
(b) 30 km h⁻¹
(c) 36.7 km h⁻¹
(d) 40 km h⁻¹
Given
three equal distances at v₁ = 20, v₂ = 30, v₃ = 60 km h⁻¹
Asked
average speed
Concept to use
Equal distances again, so it is the harmonic mean — now of three numbers.
Put the reciprocals over a common denominator immediately — here 60 — and the sum is a single fraction. The arithmetic mean would be 36.7, which is present in the options as bait.
Trap
Extending the two-speed formula 2v₁v₂/(v₁+v₂) by guesswork. Learn the reciprocal form n/Σ(1/vi) instead — it works for any number of legs.
Q5NEET-style
An athlete completes exactly one round of a circular track of radius R in time T. Her average velocity and average speed for that round are respectively
(a) 2πR/T and zero
(b) 2R/T and 2πR/T
(c) zero and zero
(d) zero and 2πR/T
Given
one complete circular lap, radius R, time T
Asked
average velocity and average speed
Concept to use
After a full lap the finish point is the start point, so the displacement is exactly zero, while the odometer has recorded the full circumference.
Formula or rule
v̄ = Δx/Δt ; average speed = 2πR/T
Baby steps
Displacement over one full lap = 0, so v̄ = 0.
Path length = 2πR.
average speed = 2πR/T.
Answer
zero and 2πR/T
Shortcut
Any closed path gives zero average velocity, whatever its shape or length. Ask ‘did it come back to where it started?’ before calculating anything.
Trap
Concluding the athlete was somehow at rest. Average velocity being zero says nothing about the instantaneous speed, which was never zero here. For half a lap the answers become 2R/(T/2) = 4R/T and 2πR/T, in the ratio 2:π.
Q6NEET-style
The position of a particle is x = 3t² + 2t (x in metres, t in seconds). Its average velocity between t = 1 s and t = 3 s is
(a) 8 m s⁻¹
(b) 11 m s⁻¹
(c) 14 m s⁻¹
(d) 20 m s⁻¹
Given
x = 3t² + 2t, interval t = 1 s to t = 3 s
Asked
average velocity over the interval
Concept to use
Average velocity over an interval needs only the two endpoint positions — it is the slope of the chord, not anything to do with the derivative.
Formula or rule
v̄ = [x(t₂) − x(t₁)]/(t₂ − t₁)
Baby steps
x(3) = 3(9) + 2(3) = 33 m.
x(1) = 3(1) + 2(1) = 5 m.
v̄ = (33 − 5)/(3 − 1) = 28/2 = 14 m s⁻¹.
Answer
14 m s⁻¹
Shortcut
For any polynomial with constant acceleration, the average velocity over an interval equals the instantaneous velocity at the mid-time. Here v = 6t + 2 at t = 2 s gives 14 m s⁻¹. ✓
Trap
Differentiating and substituting t = 3 to get 20, or t = 1 to get 8. Both are instantaneous values and both appear in the options. Average velocity is a chord; instantaneous velocity is a tangent.
Q7NEET-style
A particle moves along a straight line with velocity v = 6 − 2t (SI units). Over the interval t = 0 to t = 5 s, its average velocity and average speed are
(a) 1 m s⁻¹ and 1 m s⁻¹
(b) 2.6 m s⁻¹ and 1 m s⁻¹
(c) 1 m s⁻¹ and 2.6 m s⁻¹
(d) 5 m s⁻¹ and 13 m s⁻¹
Given
v = 6 − 2t, from t = 0 to t = 5 s
Asked
average velocity and average speed
Concept to use
The velocity changes sign at t = 3 s, so the particle reverses. Displacement is the signed area under the v–t graph; distance is the sum of the areas taken as positive.
Formula or rule
Δx = ∫v dt ; distance = ∫|v| dt
Baby steps
v = 0 at t = 3 s — the turning point.
0 to 3 s: Δx = 6(3) − 3² = +9 m.
3 to 5 s: Δx = [6(5) − 25] − 9 = 5 − 9 = −4 m.
Net displacement = 9 − 4 = 5 m; distance = 9 + 4 = 13 m.
v̄ = 5/5 = 1 m s⁻¹; average speed = 13/5 = 2.6 m s⁻¹.
Answer
1 m s⁻¹ and 2.6 m s⁻¹
Shortcut
Always find where v = 0 first. If that instant lies inside the interval, the two answers will differ; if it does not, they are equal and you have saved yourself half the work.
Trap
Integrating straight through from 0 to 5 and calling the result the distance. That gives 5 m, which is the displacement — the reversal has already been silently cancelled. Splitting at the turning point is the entire technique.
Q8NEET-style
A ball is thrown vertically upward with a speed of 20 m s⁻¹ and returns to the thrower’s hand. Taking g = 10 m s⁻², the average velocity and average speed for the complete flight are
(a) zero and 10 m s⁻¹
(b) zero and zero
(c) 10 m s⁻¹ and 20 m s⁻¹
(d) zero and 20 m s⁻¹
Given
u = 20 m s⁻¹ up, ball returns to the same point, g = 10 m s⁻²
Asked
average velocity and average speed for the whole flight
Concept to use
The ball ends where it began, so the displacement is zero. The path length is twice the maximum height.
Formula or rule
T = 2u/g ; H = u²/2g ; average speed = 2H/T
Baby steps
T = 2(20)/10 = 4 s; H = 400/20 = 20 m.
Displacement = 0, so v̄ = 0.
Path length = 2(20) = 40 m.
average speed = 40/4 = 10 m s⁻¹.
Answer
zero and 10 m s⁻¹
Shortcut
For any complete up-and-down flight the average speed is exactly u/2, because the speed falls linearly from u to 0 and back. Here 20/2 = 10. No need for H or T at all.
Trap
Answering ‘zero and zero’. Average speed can only be zero if the body never moved. Also watch the wording: for the upward half only, both answers are 10 m s⁻¹ because there is no reversal yet.
Q9NEET-style
Which of the following is always true for motion along a straight line?
(a) Average speed is always equal to the magnitude of the average velocity
(b) Average speed is always less than the magnitude of the average velocity
(c) Average speed is greater than or equal to the magnitude of the average velocity
(d) Instantaneous speed can exceed the magnitude of the instantaneous velocity
Given
general statements about average and instantaneous quantities
Asked
the statement that always holds
Concept to use
The path length between two points can never be shorter than the straight-line separation, and the two are divided by the same time. So the inequality runs one way only.
Formula or rule
path length ≥ |displacement| ⇒ average speed ≥ |v̄|
Baby steps
They are equal only when the body never reverses direction — then the path is the displacement.
The moment the body turns around, the path keeps growing while the displacement shrinks, so the average speed becomes strictly greater.
Instantaneous speed always equals the magnitude of the instantaneous velocity, because in a vanishingly small interval there is no time to double back — so the last option is false.
Answer
Average speed is greater than or equal to the magnitude of the average velocity
Shortcut
Over an interval: inequality. At an instant: equality. That single contrast answers most conceptual questions in this topic.
Trap
Picking the second option because ‘speed is the magnitude of velocity’. That is true instantaneously but not for averages, because the two averages are built from different numerators.
Q10NEET-style
A person walks 4 m east, then 3 m west, then 5 m east, taking 10 s in all. The ratio of the average speed to the magnitude of the average velocity is
(a) 1 : 1
(b) 2 : 1
(c) 6 : 5
(d) 12 : 5
Given
legs of +4 m, −3 m, +5 m; total time 10 s
Asked
average speed ÷ |average velocity|
Concept to use
The ratio is just path length ÷ |displacement| — the time cancels, so you never need to compute either average.
Formula or rule
ratio = (total path length)/|displacement|
Baby steps
Displacement = +4 − 3 + 5 = +6 m.
Path length = 4 + 3 + 5 = 12 m.
Ratio = 12/6 = 2, i.e. 2 : 1.
(For the record, the averages themselves are 1.2 and 0.6 m s⁻¹.)
Answer
2 : 1
Shortcut
Whenever a question asks for the ratio of these two, ignore the time entirely. It appears in both denominators and cancels.
Trap
Adding the leg lengths with signs to get the path (giving 6 and a ratio of 1:1), or adding them without signs to get the displacement (giving 12 and a ratio of 1:1 again). Path ignores sign; displacement respects it.
Q11NEET-style
A body moving with uniform acceleration has a velocity of 4 m s⁻¹ at one instant and 16 m s⁻¹ a while later. Its average velocity over that interval is
(a) 8 m s⁻¹
(b) 12 m s⁻¹
(c) 20 m s⁻¹
(d) 10 m s⁻¹
Given
u = 4 m s⁻¹, v = 16 m s⁻¹, uniform acceleration
Asked
average velocity
Concept to use
For uniform acceleration the velocity changes linearly with time, so its time-average is simply the midpoint of the two end values.
Formula or rule
v̄ = (u + v)/2 — uniform acceleration only
Baby steps
v̄ = (4 + 16)/2 = 10 m s⁻¹.
Equivalently, on the v–t graph the shape is a trapezium and its mean height is the average of the parallel sides.
Answer
10 m s⁻¹
Shortcut
The mid-time velocity is (u+v)/2 = 10. Note that the mid-point velocity would be √[(16 + 256)/2] = 11.66 m s⁻¹ — larger, always.
Trap
Applying (u+v)/2 when the acceleration is not uniform. That formula is a special case, not a definition; the definition is always Δx/Δt. NEET sets non-uniform versions to punish the habit.
Q12NEET-style
A student states: ‘Since the average velocity of my journey was zero, I must have been at rest the whole time.’ This conclusion is
(a) correct, because zero velocity means no motion
(b) correct only if the journey was along a straight line
(c) incorrect, because average velocity is always zero
(d) incorrect, because the body may have returned to its starting point
Given
the claim that zero average velocity implies rest
Asked
whether the reasoning holds
Concept to use
Average velocity depends only on the two endpoints. Any journey that ends where it began has zero average velocity, however energetic it was in between.
Formula or rule
v̄ = (xfinal − xinitial)/Δt
Baby steps
If xfinal = xinitial then v̄ = 0, regardless of the path.
A body genuinely at rest also has zero average velocity — so the condition does not distinguish the two cases.
The average speed is what would have been zero for a body truly at rest.
Answer
incorrect, because the body may have returned to its starting point
Shortcut
Average velocity throws away everything except the endpoints. To detect whether motion occurred at all, you need the average speed or the instantaneous velocity.
Trap
This appears as an assertion–reason item: assertion true, reason true, but the reason does not explain the assertion. Read carefully which of the two quantities each half refers to.
Q13NEET-style
A train travels the first 30 km of a 60 km trip at 30 km h⁻¹. To achieve an average speed of 60 km h⁻¹ for the whole trip, it must cover the remaining 30 km at
(a) no speed can achieve this
(b) 60 km h⁻¹
(c) 90 km h⁻¹
(d) 120 km h⁻¹
Given
d₁ = d₂ = 30 km, v₁ = 30 km h⁻¹, target average = 60 km h⁻¹
Asked
required speed for the second half
Concept to use
Work with time, not speed. Fix the total time the target average allows, then see how much of it is already gone.
Formula or rule
ttotal = dtotal/v̄
Baby steps
Target: 60 km at 60 km h⁻¹ means the whole trip must take 1 h.
The first 30 km already took 30/30 = 1 h.
That leaves 0 h for the remaining 30 km.
No finite speed covers 30 km in zero time — the target is unreachable.
Answer
no speed can achieve this
Shortcut
Whenever the first leg is run at exactly half the target average over half the distance, the whole time budget is spent. Check the time budget before hunting for a speed.
Trap
Answering 90, by averaging 30 and 90 to get 60. Arithmetic averaging of speeds is wrong here for the usual reason: the legs take different times. This question exists purely to punish that reflex.
Q14NEET-style
The magnitude of the average velocity of a particle equals its average speed over some interval. This means that during the interval the particle
(a) did not reverse its direction of motion
(b) was at rest
(c) moved with uniform velocity
(d) had zero acceleration
Given
|average velocity| = average speed over an interval
Asked
what this implies about the motion
Concept to use
Equality holds exactly when the path length equals the magnitude of the displacement — that is, when the motion never doubles back.
Formula or rule
|Δx| = path length ⇔ no reversal
Baby steps
If the particle turns around, some path is retraced and the path length exceeds |Δx|, breaking the equality.
If it never turns around, every metre of path adds a metre of displacement in the same direction, so they are equal.
The particle may still speed up and slow down freely, so uniform velocity and zero acceleration are sufficient but not necessary.
Answer
did not reverse its direction of motion
Shortcut
The condition constrains the sign of the velocity, not its size. One-way motion — however erratic in magnitude — is exactly what is required.
Trap
Choosing ‘uniform velocity’. That would certainly give equality, but so would any accelerating one-way journey. NEET rewards the weakest sufficient condition, not the most familiar one.
Q15NEET-style
A particle covers half the total distance with speed v₁. The remaining half is covered in two equal time intervals with speeds v₂ and v₃. The average speed over the whole journey is
(a) (v₁ + v₂ + v₃)/3
(b) 2v₁(v₂ + v₃)/(2v₁ + v₂ + v₃)
(c) (v₂ + v₃)/2
(d) 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁)
Given
first half distance at v₁; second half covered in two equal times at v₂ and v₃
Asked
average speed for the whole journey
Concept to use
Handle each half with its own rule: equal distance uses the given speed directly, equal times give an arithmetic mean for that half. Then combine the two halves as equal distances.
Formula or rule
second-half average = (v₂ + v₃)/2, then v̄ = 2ab/(a + b)
Baby steps
Second half: equal times, so its average speed is a = (v₂ + v₃)/2.
Now the journey is two equal distances at v₁ and a.
v̄ = 2v₁a/(v₁ + a).
Substitute a: v̄ = 2v₁(v₂+v₃)/(2v₁ + v₂ + v₃).
Answer
2v₁(v₂ + v₃)/(2v₁ + v₂ + v₃)
Shortcut
Sanity-check any symbolic answer by setting all three speeds equal to v: this expression gives 2v(2v)/(4v) = v. ✓ Options that fail this test can be struck out immediately.
Trap
Applying one rule to the whole journey. This is a deliberately mixed question — equal distances in the outer structure, equal times in the inner. Deal with the inner structure first.
Q16NEET-style
The average speed of a body is 20 m s⁻¹ over a certain interval. Which of the following can be concluded about its instantaneous speed?
(a) It was 20 m s⁻¹ throughout
(b) It reached 20 m s⁻¹ at least once during the interval
(c) It never exceeded 20 m s⁻¹
(d) It was zero at the start and end of the interval
Given
average speed = 20 m s⁻¹ over an interval
Asked
what follows about the instantaneous speed
Concept to use
An average must lie between the minimum and maximum of the quantity being averaged. For a speed that varies continuously, that forces it to take the average value at some instant.
Formula or rule
mean value property: a continuous function attains its mean on the interval
Baby steps
If the speed were always below 20, the average would be below 20. Same argument above.
So the speed must be at or above 20 somewhere and at or below 20 somewhere.
Being continuous, it must pass through exactly 20 at least once in between.
It may well have exceeded 20 at other moments, so the third option fails.
Answer
It reached 20 m s⁻¹ at least once during the interval
Shortcut
An average is always ‘visited’ by a continuously changing quantity. The same reasoning underlies why the average velocity of a uniformly accelerated body equals its mid-time velocity.
Trap
Assuming an average describes the whole interval. It constrains the interval as a whole; it says very little about any particular instant except this one existence result.
Answer key
aQ1
cQ2
dQ3
bQ4
dQ5
cQ6
cQ7
aQ8
cQ9
bQ10
dQ11
dQ12
aQ13
aQ14
bQ15
bQ16
NEET 2027 preparation — built for Aamirah Fathima. Source: NCERT Physics Class XI,
Chapter 2 (Reprint 2026–27). Single self-contained file — print directly from any browser.