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NEET 2027 · Chapter 2 · Motion in a Straight Line

Every piece of maths this chapter asks of you

Nineteen small ideas, each explained plainly, drawn as a moving diagram, and tied to the exact questions where it is needed.
19 topics95 examplesToolkit19 animated diagramsClass 6 to Class 11

How to use this page

The one idea

Chapter 2 is not hard physics. It is easy physics resting on nineteen small pieces of school mathematics. When a question feels impossible, it is almost always one of these nineteen that has gone wrong — not the physics.

Each section below has four parts: an explanation written so a ten-year-old could follow it, a diagram that moves, a list of the exact places in Chapter 2 where the idea is needed, and five worked examples pointing at real questions in Files 1–7.

Nothing here is beyond Class 10 except the last five sections, which are the Class 11 calculus that NCERT introduces inside this very chapter. Even those reduce to one small rule each.

If you only fix three things, fix these

1. The × 5/18 conversion (section 02) — skipping it produces absurd answers, and it opens more questions in this chapter than any other single step.
2. The sign convention (section 01) — choose your plus direction once and never change it.
3. Slope and area (sections 10 and 11) — between them they answer every graph question in the chapter without any algebra at all.

01 · Class 6 arithmeticPlus and minus — the sign convention02 · Class 6 fractionsFractions, and the one conversion you cannot skip03 · Class 6–7 ratioRatio and proportion04 · Class 7–8Squares, square roots and surds05 · Class 7–8 algebraRearranging a formula06 · Class 7 algebraSolving a simple (linear) equation07 · Class 10 algebraQuadratic equations — and why two answers is normal08 · Class 9–10 algebraTwo equations at once (simultaneous equations)09 · Class 6–8Coordinates — reading a point off a graph10 · Class 9–10Slope — how steep is it?11 · Class 6–8 mensurationArea of a rectangle, triangle and trapezium12 · Class 9–10Straight lines and parabolas — reading a shape13 · Class 8–10Sequences — and Galileo’s odd numbers14 · Class 8–10Three kinds of average15 · Class 11 calculusDifferentiation — the power rule16 · Class 11 calculusIntegration — the power rule backwards17 · Class 11, but the idea is simpleLimits — squeezing the gap to nothing18 · Class 11Exponentials and logarithms19 · Class 6–7, plus judgementRounding, decimals and how many figures to keep
01

Plus and minus — the sign convention

Class 6 arithmetic
In plain words — the ten-year-old version

Imagine standing on a footpath. You decide, before you start, that walking towards the shop counts as plus and walking home counts as minus. Now ‘−3 steps’ is not a strange number — it just means three steps the other way.

The only rule is: choose which way is plus once, and never change your mind halfway. That is the entire secret. Almost every lost mark in this chapter is somebody who changed their mind.

Two straight lines around a number, like |−20|, mean ‘forget the sign, just how big is it’. So |−20| = 20. That is the difference between velocity (has a sign) and speed (never does).

-3-2-10+1+2+3 this way counts as PLUS this way counts as MINUS Same speed, opposite signs: +20 m/s going up, −20 m/s coming down.
animated Pick your plus direction first. Then a minus sign is just a direction, not something scary.
Where you actually need it
  • Throwing a ball up: u = +20, a = −10. The signs disagree, and that is exactly what ‘slowing down’ looks like in algebra.
  • Retardation is written as a negative acceleration. ‘Retardation 3 m s⁻²’ means a = −3 in your working.
  • Velocity can be negative; speed and distance never can.
  • Relative velocity: vAB = −vBA — same size, opposite sign.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
+20 m/s and −20 m/ssame speed, opposite waysFile 3 — a ball leaves your hand and returns at the same speed reversed
a = −10 m s⁻² going up
a = −10 m s⁻² coming down
unchangedFile 3 — gravity points down the whole flight, whichever way the ball moves
displacement −4 m
distance +4 m
over t = 3 s to 5 sFile 6 Q7 — where the two averages come apart
|−9| = 9File 7 Q3 — the speed is 9 m/s, the velocity is −9 m/s
vAB = +12, so vBA= −12File 5 Q14 — swapping the two bodies flips the sign only
02

Fractions, and the one conversion you cannot skip

Class 6 fractions
In plain words — the ten-year-old version

A fraction is just a sharing instruction. 5/18 means ‘cut it into 18 equal bits and keep 5 of them’.

Speeds in exam questions arrive in kilometres per hour, but physics formulas only work in metres per second. To swap, multiply by 5/18. It is not a magic number: a kilometre is 1000 metres and an hour is 3600 seconds, and 1000/3600 tidies up to 5/18.

Quick sanity check: the answer in m/s should always be smaller than the number in km/h, roughly a third of it. If your answer got bigger, you multiplied by 18/5 by mistake.

THE ONLY MACHINE × 5/18 72 km/h 20 m/s Why 5/18? One kilometre is 1000 metres, one hour is 3600 seconds. 1000 ÷ 3600 = 10/36 = 5/18
animated One kilometre per hour is a slow crawl. The 5/18 machine shrinks the number by roughly two-thirds.
Where you actually need it
  • Every stopping-distance and braking question opens with a speed in km h⁻¹, on purpose.
  • NCERT Exercise 2.5 (126 km h⁻¹) and Exercise 2.14 (36 and 54 km h⁻¹) both start with this step.
  • Train-crossing questions in File 5 give both trains in km h⁻¹.
  • If you forget it, your acceleration comes out around 40 m s⁻² — four times gravity, which no car can do. That absurdity is your warning bell.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
1000 ÷ 3600= 5/18the reason the factor exists at all — worth deriving once
18 × 5/18= 5 m/sthe easiest one to check the rule with
126 × 5/18= 35 m/sFile 2 Q1 — NCERT Exercise 2.5, the braking car
54 × 5/18= 15 m/sFile 5 Q1 — NCERT Exercise 2.14, cars B and C
192 × 5/18= 53.3 m/sFile 5 Q16 — the thief’s speeding car
03

Ratio and proportion

Class 6–7 ratio
In plain words — the ten-year-old version

A ratio compares two things: 1 : 3 means the second one is three times the first. You can multiply both sides by anything and the ratio does not change — 1 : 3 is the same as 2 : 6.

‘Proportional to’ (written ) means: if one doubles, so does the other. But watch for proportional to the square. If the stopping distance goes as , then doubling your speed does not double the road you need — it needs four times as much. Triple the speed and you need nine times.

That single fact is why school-zone speed limits exist, and it is examined every year.

SPEED v 1 car length of road SPEED 2v — only twice as fast 4 car lengths of road Double the speed, and the road you need is not doubled — it is squared.
animated A car at 40 km/h needs one car-length to stop. The same car at 80 km/h needs four. This is why the extra speed is so dangerous.
Where you actually need it
  • Stopping distance ∝ v², but stopping time ∝ v — two different powers in the same question.
  • Free fall from rest: distances after 1, 2, 3, 4 s are in the ratio 1 : 4 : 9 : 16.
  • Distances in successive seconds are in the ratio 1 : 3 : 5 : 7 — Galileo’s odd numbers.
  • Ratio questions in File 6 ask for average speed ÷ |average velocity|, where the time cancels and only the ratio survives.
  • Watch the order. Writing the ratio upside down is one of the recorded recurring slips — read which quantity the question names first.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
distances after 1, 2, 3, 4 s from rest1 : 4 : 9 : 16File 2 — because s ∝ t²
distances in the 1st, 2nd, 3rd, 4th second1 : 3 : 5 : 7File 2 — Galileo’s law of odd numbers
speed doubles, stopping distance× 4File 2 — NCERT braking data 10, 20, 34, 50 m
speed doubles, stopping time× 2File 2 — the contrast with the line above is the whole question
path 12 m, displacement 6 mratio 2 : 1File 6 Q10 — the time cancels, so never compute the averages
04

Squares, square roots and surds

Class 7–8
In plain words — the ten-year-old version

means ‘5 rows of 5’ — draw it and you literally get a square with 25 little boxes in it. The square root asks the question backwards: √25 means ‘25 boxes are arranged in a square — how long is one side?’ Answer: 5.

Most roots are not tidy. √40 is not a whole number; it sits between √36 = 6 and √49 = 7, at about 6.32. Leaving it as 2√10 is called a surd and is often the neater answer in an options list.

Handy trick: √40 = √(4×10) = 2√10. Pull out any square factor you can spot.

5 across 5 down 5² = 25 little squares √25 = 5 — the side Squaring builds the square. Rooting asks: “what side would give me this many?” √40 is not a whole number — it sits between 6 and 7, at 6.32.
animated 5² builds the square; √25 finds its side. Every v² = u² + 2as question is one of these two moves.
Where you actually need it
  • v² = u² + 2as — the most-used equation in the chapter is built entirely on squares.
  • Finding v from v² always needs a square root, and always at the very last step.
  • Free-fall time t = √(2h/g).
  • Velocity at the mid-point of a journey is a root-mean-square: √[(u²+v²)/2].
  • NEET options are often left in surd form, so 2√10 and √40 may both appear — they are the same number.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
35²= 1225File 2 Q1 — the braking car, before dividing by 400
5² − 4²= 25 − 16 = 9 mFile 2 Q2 — why the distance in the 5th second is 9 m
√(2 × 2 ÷ 12)= 0.577 sFile 5 Q10 — the coin dropped in a lift
√40= 2√10 = 6.32 m/sFile 7 Q14 — speed from the area under an a–x graph
√250= 15.81 m/sFile 2 — a v² = u² + 2as answer that stays untidy
05

Rearranging a formula

Class 7–8 algebra
In plain words — the ten-year-old version

A formula is a see-saw that is perfectly balanced. You are allowed to change it, on one condition: whatever you do to the left, do exactly the same to the right. Take 3 off one side, take 3 off the other, and it stays level.

So to get a out of v² = u² + 2as: take off both sides, then divide both sides by 2s. Two moves, and a is alone.

People call this ‘taking it to the other side and changing the sign’. That works, but it is a shortcut for the see-saw idea — and when a question gets messy the see-saw picture is the one that saves you.

u² + 2as − u² − u² Whatever you do to one side, do to the other — and it stays level. v² − u² = 2as  →  a = (v² − u²) / 2s
animated Take the same thing off both pans and the balance is undisturbed. That is all rearranging is.
Where you actually need it
  • You will rearrange v² = u² + 2as for a, for s and for v — all three appear in File 2.
  • s = ut + ½at² rearranged for a is the last step of NCERT Exercise 2.14.
  • H = u²/2g rearranged for u answers ‘how fast must I throw it to reach this height?’
  • The escalator formula 1/t = 1/t₁ + 1/t₂ has to be flipped at the end — a favourite place to stop one step too early.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
v² = u² + 2as, make a the subjecta = (v² − u²)/2sFile 2 Q1 — finding the retardation of the braking car
s = ut + ½at², make a the subjecta = 2(s − ut)/t²File 5 Q1 — the minimum acceleration in NCERT Exercise 2.14
v = u + at, make t the subjectt = (v − u)/aFile 2 — every ‘how long to stop?’ question
H = u²/2g, make u the subjectu = √(2gH)File 3 — the launch speed needed for a given height
1/t = 1/t₁ + 1/t₂, make t the subjectt = t₁t₂/(t₁ + t₂)File 5 Q6 — the escalator; you must flip at the end
06

Solving a simple (linear) equation

Class 7 algebra
In plain words — the ten-year-old version

An equation with one t in it and no is called linear, and it always has exactly one answer. Think of it as a parcel wrapped in layers: unwrap them in the reverse of the order they went on.

0 = 35 − 3.06t. The t has been multiplied by 3.06 and then taken away from 35. So undo the subtraction first (move it across), then undo the multiplication (divide). Two moves.

Always check by putting your answer back in: 35 − 3.06 × 11.4 really is about 0. Ten seconds of checking has saved many marks.

0 = 35 − 3.06 t the car stops, so the final velocity is 0 3.06 t = 35 move the t-term across — add 3.06 t to both sides t = 35 ÷ 3.06 = 11.4 s divide both sides — done
animated Unwrap in reverse order. Move the term across, then divide. Then check by substituting back.
Where you actually need it
  • Finding the stopping time once you know the retardation.
  • Finding the instant when two moving bodies are at the same place, once the terms have cancelled.
  • Finding when acceleration is zero, which is where velocity is largest.
  • Finding when velocity is zero, which is where the body turns round — the essential first step in every distance-versus-displacement question.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
0 = 35 − 3.0625 tt = 11.4 sFile 2 Q1 — how long the car takes to stop
1000 = 200 + 800 aa = 1 m s⁻²File 5 Q1 — NCERT Exercise 2.14
20t − 5t² = 20(t−2) − 5(t−2)²t = 3 sFile 5 Q12 — the t² terms cancel, leaving a linear equation
2 − 0.5 t = 0t = 4 sFile 7 Q11 — where the velocity is greatest
6 − 2 t = 0t = 3 sFile 6 Q7 — where the particle turns round
07

Quadratic equations — and why two answers is normal

Class 10 algebra
In plain words — the ten-year-old version

An equation with a in it is a quadratic, and it usually has two answers. That feels wrong until you see why.

Throw a ball straight up. Ask ‘when was it 40 m high?’ The honest answer is twice — once climbing and once falling. The two answers are not a mistake; they are two real moments.

Two ways to solve one. If it factorises, do that: t² − 6t + 8 = (t − 2)(t − 4), so t = 2 or 4. If it does not, use the formula t = [−b ± √(b² − 4ac)]/2a.

Then think. Sometimes one answer is nonsense — a negative time means ‘before you threw it’. Throw that root away and say why.

40 m t = 2 s t = 4 s Thrown up at 30 m/s, the ball is 40 m high TWICE — on the way up and on the way down. time height
animated Two answers, two real moments. But a negative time means ‘before the story started’ — reject it and say so.
Where you actually need it
  • Every ‘when does it reach height h?’ question in File 3 is a quadratic with two roots.
  • Chase problems: t² − 10t − 75 = 0 gives t = 15 and t = −5; only the positive one is a real catch.
  • Bodies thrown from a height give roots such as 5 and −1 — discard the negative.
  • A body returning to its starting point gives t = 0 as one root. That is the moment it set off, so reject it and keep the other.
  • If the discriminant b² − 4ac is negative there is no real answer — which means the ball never reaches that height at all.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
5t² − 30t + 40 = 0t = 2 s and t = 4 sFile 3 — the ball is at 40 m twice
t² − 10t − 75 = 0t = 15 s (reject −5)File 5 Q8 — the police jeep catching the thief
5t² − 20t − 25 = 0t = 5 s (reject −1)File 3 — NCERT Example 2.3, the ball from the rooftop
t(2t − 10) = 0t = 0 or t = 5 sFile 5 Q13 — t = 0 is the start, so the meeting is at 5 s
t(10 − t²/3) = 0t = √30 = 5.48 sFile 7 Q15 — returning to the origin under a = −2t
08

Two equations at once (simultaneous equations)

Class 9–10 algebra
In plain words — the ten-year-old version

When two things are moving, each has its own rule for where it is. Car P follows x = 20 + 15t. Car Q follows x = 140 + 3t.

‘When do they meet?’ means: when do the two rules give the same answer? So set them equal and solve. On a graph, that is simply where the two lines cross.

The big warning: meeting means same position, not same speed. When a police jeep matches the thief’s speed it is still behind him — that is just the moment the gap stops growing. Papers put both answers in the options every single time.

t = 10 s they meet here P: 20 + 15t Q: 140 + 3t Two rules, one moment when they agree. That crossing point is the answer. time position
animated Set the two position rules equal. Where the lines cross is where the bodies meet.
Where you actually need it
  • Every overtaking, chasing and catching-up question in File 5.
  • Two stones thrown at different times meeting in mid-air.
  • Two particles starting together with different accelerations meeting again.
  • Reading the crossing point straight off an x–t graph in File 4 — no algebra needed at all.
  • Choosing a frame (sitting in one of the cars) turns two equations into one, which is why File 5 keeps recommending it.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
20 + 15t = 140 + 3tt = 10 s, x = 170 mFile 5 Q11 — P overtakes Q
t² = 75 + 10tt = 15 sFile 5 Q8 — jeep from rest vs thief at steady speed
20t − 5t² = 20(t−2) − 5(t−2)²t = 3 s, height 15 mFile 5 Q12 — two stones thrown 2 s apart
½(4)t² = 10tt = 5 sFile 5 Q13 — accelerating body catches a steady one
50t + 50t = 100 kmt = 1 hFile 5 Q9 — two cars approaching, which fixes how long the bird flies
09

Coordinates — reading a point off a graph

Class 6–8
In plain words — the ten-year-old version

A graph is a map. To find a place on it you always do the same two moves: across first, then up. That is why a point is written (4, 1) — across 4, up 1.

In this chapter the across direction is nearly always time, and the up direction is position, velocity or acceleration depending on the graph. So the very first thing to do with any graph is read the two axis labels. Two graphs can look identical and mean completely different things.

Where a curve crosses the across-axis, the up-value is zero. On a v–t graph that means the body has stopped — usually the most important instant in the whole question.

12 34 5 12 1. go ACROSS — this is the time 2. then go UP — this is the position (4 s, 1 m) Always across first, then up. Every graph in this chapter is read this way.
animated Across, then up. And read the axis labels before anything else — they change what the picture means.
Where you actually need it
  • Reading NCERT Table 2.1 as points on the x = 0.08t³ curve.
  • Every question in File 4 begins by reading two points off a line.
  • The starting position x₀ is the value where the line meets the up-axis (time zero).
  • Where a v–t line crosses the time axis, the body is momentarily at rest and about to reverse.
  • Where two x–t lines cross, the two bodies are at the same place — they have met.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
read x at t = 4 s on x = 0.08t³x = 5.12 mFile 1 — NCERT Table 2.1
where an x–t line meets the up-axisthat value is x₀File 4 — the starting position, often not zero
where a v–t line crosses the time axisv = 0, the turning pointFile 6 Q7 — split the journey here
where two x–t lines crosssame position → they meetFile 4 & File 5 Q11
axis says ‘x’ or axis says ‘v’completely different meaningFile 4 — the most common misread in the chapter
10

Slope — how steep is it?

Class 9–10
In plain words — the ten-year-old version

Slope means steepness, and you measure it the same way you would measure a staircase: how far up, divided by how far along. Rise over run. A gentle ramp has a small slope; a steep one has a big slope. Going downhill gives a negative slope.

This is the single most useful idea in the whole chapter, because on a graph the slope is a physical quantity:

• slope of a position–time graph = velocity
• slope of a velocity–time graph = acceleration

On a straight line the slope is the same everywhere, so pick any two points. On a curve it keeps changing, so you slope the tangent — the straight line that just kisses the curve at that one spot.

run rise slope = rise ÷ run STRAIGHT LINE — same steepness everywhere CURVE — steepness changes so you slope the TANGENT instead
animated Rise over run. On a straight line, any two points will do. On a curve, use the tangent at the point you care about.
Where you actually need it
  • Velocity from an x–t graph, and acceleration from a v–t graph — the backbone of File 4.
  • Chord slope gives the average velocity over an interval; tangent slope gives the instantaneous velocity at a point. That contrast is the whole of File 1.
  • A negative slope means the body is moving backwards.
  • A horizontal line (zero slope) on an x–t graph means the body is standing still.
  • Relative velocity is the difference of two slopes on the same x–t graph.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
(33 − 5) ÷ (3 − 1)= 14 m/sFile 6 Q6 — chord slope, so this is the average velocity
tangent slope at t = 4 s on x = 0.08t³= 3.84 m/sFile 1 — tangent slope, so this is the instantaneous velocity
slope of a v–t line= accelerationFile 4 — a straight v–t line means constant acceleration
slope of an x–t line = 0the body is at restFile 4 — a flat line is not ‘no graph’, it is standing still
15 − 3= 12 m/s relativeFile 5 Q11 — relative velocity is the gap between two slopes
11

Area of a rectangle, triangle and trapezium

Class 6–8 mensuration
In plain words — the ten-year-old version

Three shapes, three formulas you already know:

• rectangle = length × width
• triangle = ½ × base × height
• trapezium = ½ × (top + bottom) × height

Here is why they matter. On a velocity–time graph, the area underneath the line is the distance travelled. If you go 6 m/s for 4 s you cover 24 m — and that is exactly the area of a 6-by-4 rectangle. The maths and the physics are the same picture.

If the line slopes, split the shape into a rectangle plus a triangle, or use the trapezium formula in one go. Both give the same answer.

One catch: area below the time axis counts as negative, because the body is going backwards. That is how displacement and distance come apart.

6 18 4 s rectangle = 6 × 4 = 24 m triangle = ½ × 4 × 12 = 24 m total 48 m = how far it went On a velocity–time graph, the AREA underneath is the distance travelled. Area below the time axis counts as NEGATIVE — that is the body going backwards.
animated Area under a v–t graph is distance. Split into a rectangle plus a triangle, or use the trapezium formula.
Where you actually need it
  • The whole derivation of x = v₀t + ½at² is one rectangle plus one triangle — NCERT Fig. 2.5.
  • Every ‘find the displacement from the graph’ question in File 4.
  • Distance versus displacement: add the areas with signs for displacement, without signs for distance.
  • Area under an a–t graph is the change in velocity, not distance. Different graph, different meaning — check the axis.
  • Area under an a–x graph is the change in v²/2. A third meaning again.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
6 m/s held for 4 srectangle = 24 mFile 2 — the v₀t part of the second equation
½ × 4 × 12triangle = 24 mFile 2 — the ½at² part, the bonus the acceleration buys
½ × (6 + 18) × 4trapezium = 48 mFile 4 Q17 — both halves in one step
areas +9 m and −4 mdisplacement 5 m, distance 13 mFile 6 Q7 — the reversal splits the two answers
area under a–t of 2 m/s² for 2 sΔv = 4 m/sFile 4 — a–t area gives velocity, not distance
12

Straight lines and parabolas — reading a shape

Class 9–10
In plain words — the ten-year-old version

You can tell what a body is doing just from the shape of its graph, without a single calculation.

If something has t in it but no , its graph is a straight line. That is what steady speed looks like: equal steps in equal times.

If it has a in it, the graph is a parabola — a curve that gets steeper and steeper. That is what speeding up looks like. If the has a minus in front, the parabola opens downwards, like the path of a thrown ball.

So: x = v₀t + ½at² has a in it, which is exactly why every constant-acceleration x–t graph is a curve, never a line.

STEADY SPEED equal steps in equal times a straight line SPEEDING UP bigger and bigger steps a parabola (a curve)
animated No t² means a straight line. A t² means a curve. That one glance answers most ‘which graph?’ questions.
Where you actually need it
  • ‘Which graph shows uniform motion?’ — x–t straight, v–t flat. File 4 Q7.
  • ‘Which graph shows uniform acceleration?’ — x–t parabola, v–t straight, a–t flat. Each is one rung simpler than the one before.
  • A downward-opening parabola is the signature of a body thrown upward.
  • Impossible graphs: a position–time graph that goes straight up would mean being in two places at once. NCERT Exercise 2.12.
  • A sharp corner on a v–t graph would mean infinite acceleration — also impossible.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
x = 20 + 15tstraight lineFile 5 Q11 — no t², so steady speed
x = v₀t + ½at²parabolaFile 2 — the t² is what makes it bend
v = v₀ + atstraight lineFile 4 — velocity climbs steadily under constant acceleration
a = −9.8 in free fallflat horizontal lineFile 3 — the a–t graph is the simplest of the three
y = 25 + 20t − 5t²downward parabolaFile 3 — NCERT Example 2.3, the ball from the rooftop
13

Sequences — and Galileo’s odd numbers

Class 8–10
In plain words — the ten-year-old version

A sequence is just a list of numbers with a pattern. In an arithmetic progression you add the same amount each time: 1, 3, 5, 7 (add 2 each step).

Here is a beautiful fact about falling. Drop a ball. After 1, 2, 3, 4 seconds it has fallen distances in the ratio 1 : 4 : 9 : 16 — the square numbers. Now look at the gaps between those: 4 − 1 = 3, 9 − 4 = 5, 16 − 9 = 7. So in each successive second it falls 1, 3, 5, 7 — the odd numbers.

Galileo spotted this in about 1604, and it is still on the syllabus. It works because the gap between consecutive square numbers is always the next odd number.

The trap: ‘distance in 5 seconds’ is a total. ‘Distance in the 5th second’ is one slice. They are not the same number and both will be in the options.

1 mafter 1 s4 mafter 2 s9 mafter 3 s16 mafter 4 s1 m in second 13 m in second 25 m in second 37 m in second 4 Totals go 1, 4, 9, 16 — but the GAPS go 1, 3, 5, 7. The odd numbers.
animated Totals are the squares; the gaps between them are the odd numbers. That is Galileo’s law of odd numbers.
Where you actually need it
  • Sn = u + (a/2)(2n − 1) — the distance in the nth second, which is an arithmetic progression with common difference a.
  • Free-fall questions asking for the ratio of distances in successive seconds — the answer is always odd numbers when starting from rest.
  • Reversing time: the distance covered in the last second before stopping is always a/2, whatever the starting speed.
  • Any question where the phrase ‘in the nth second’ appears.
  • Assertion–reason items sometimes attach the law of odd numbers to the wrong scientist — it is Galileo.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
distances after 1, 2, 3, 4 s1 : 4 : 9 : 16File 2 — the totals, because s ∝ t²
gaps between those totals1 : 3 : 5 : 7File 2 — Galileo’s odd numbers
S₅ with u = 0, a = 2= (2/2)(2×5 − 1) = 9 mFile 2 Q2 — the 5th second, not the first five seconds
check it: 5² − 4²= 25 − 16 = 9 mFile 2 Q2 — the subtraction check, always worth doing
distance in the last second before stopping= a/2, whatever u wasFile 2 — comes from running the motion backwards
14

Three kinds of average

Class 8–10
In plain words — the ten-year-old version

‘Average’ sounds like one thing. It is three, and the exam sets them against each other.

Arithmetic mean — add and divide. Use it when each speed was held for the same amount of time.

Harmonic mean2v₁v₂/(v₁+v₂). Use it when each speed covered the same distance. Why is it smaller? Because you spend longer crawling at the slow speed, so the slow speed counts for more.

Root-mean-square — square them, average, square-root. This is what turns up at the half-way point of a journey.

For 40 and 60 the three answers are 48, 50 and 51. They sit close together deliberately, so all three will be in the options. The words distance or time in the question tell you which one to pick.

48.0harmonic meanequal distances50.0arithmetic meanequal times50.99root-mean- squaremid-point speed Same two speeds — 40 and 60 km/h — three different “averages”. The words in the question decide which one is correct. They are always close together, on purpose.
animated Equal distances → harmonic. Equal times → arithmetic. Half-way point → root-mean-square.
Where you actually need it
  • The two-speed journey question, which appears in almost every mock paper.
  • ‘Half the journey’ versus ‘half the time’ — one word changes the formula and the answer.
  • The mid-time velocity (u+v)/2 versus the mid-point velocity √[(u²+v²)/2].
  • Round-trip boat problems, where the harmonic mean is always below the still-water speed.
  • Average speed is never just the average of the speeds unless the times were equal — this is the single most common slip in File 6.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
equal times at 40 and 60(40+60)/2 = 50 km/hFile 6 Q3 — arithmetic mean
equal distances at 40 and 602(40)(60)/100 = 48 km/hFile 6 Q2 — harmonic mean, always the smaller
equal distances at 20, 30, 603 ÷ (1/20+1/30+1/60) = 30File 6 Q4 — harmonic mean of three
mid-TIME velocity, u = 4, v = 16(4+16)/2 = 10 m/sFile 6 Q11 — arithmetic
mid-POINT velocity, u = 4, v = 16√[(16+256)/2] = 11.66 m/sFile 2 — root-mean-square, always the larger
15

Differentiation — the power rule

Class 11 calculus
In plain words — the ten-year-old version

Differentiating sounds frightening. For everything in this chapter it is one small trick.

Bring the little number down to the front, then take one off it. So becomes 3t². And becomes 2t. And t (which is secretly ) becomes 1.

A plain number on its own becomes zero, because a number never changes and differentiating asks ‘how fast is this changing?’

What it means: differentiating position gives velocity, and differentiating velocity gives acceleration. It is the slope of the tangent, worked out by algebra instead of by drawing.

Do it in pieces and add them up — each term separately, signs kept.

t 3 1. the power hops to the front becomes 3t 2 2. and what is left drops by one Same rule every time. A plain number just disappears — a number never changes. t⁵→5t⁴   6t→6   4→0   t³−6t²+3t+4 → 3t²−12t+3
animated Power to the front, then knock one off. Constants vanish. Do each term separately and add.
Where you actually need it
  • Going down the ladder: position → velocity → acceleration. File 1 is built on this.
  • Finding when the velocity is zero (the turning point) from a given x(t).
  • Finding when the acceleration is zero, which is where the velocity is largest or smallest.
  • Checking whether a given motion has constant acceleration — differentiate twice and see if the answer still has a t in it.
  • Every previous-year question of the form ‘x = …, find v or a at t = …’.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
x = 0.08t³v = 0.24t²File 1 — NCERT Table 2.1; at t = 4 this gives 3.84 m/s
x = 8 + 12t − t³v = 12 − 3t², a = −6tFile 1 — an AIPMT item; the 8 disappears
x = 3t² + 2tv = 6t + 2File 6 Q6 — used to check the chord answer of 14 m/s
x = t³ − 6t² + 3t + 4v = 3t² − 12t + 3
a = 6t − 12
File 7 Q3 — a = 0 at t = 2 s
differentiate the number 4= 0File 7 — a constant starting position never affects the velocity
16

Integration — the power rule backwards

Class 11 calculus
In plain words — the ten-year-old version

Integration is differentiation in reverse. Instead of bringing the power down, you raise the power by one and divide by the new power. So 3t² goes back to .

There is one extra thing. When you differentiated, any plain number vanished — so going backwards you cannot tell what it was. You write + C for it and then work it out from the question: C is whatever the quantity was at the start, usually the initial velocity or the initial position.

What it means physically: integrating acceleration gives velocity, and integrating velocity gives displacement. It is the area under the graph, worked out by algebra instead of by counting squares.

Easier trick: put the limits in straight away (0 to 2, say) and the C cancels itself out.

DIFFERENTIATE — going down t³ → 3t² INTEGRATE — going back up 3t² → t³ Raise the power by one, then divide by the new power. 3t² → t³ + C C is whatever it was at the start — the initial velocity or position.
animated Raise the power, divide by the new power, add C — and C is the starting value.
Where you actually need it
  • Whenever the acceleration is not constant, the three kinematic equations are unusable and integration is the only route. All of File 7.
  • Climbing back up the ladder: acceleration → velocity → position.
  • Finding displacement from a given v(t).
  • Deriving v = u + at yourself, by integrating a constant a — which shows where the kinematic equations came from in the first place.
  • The position form ∫v dv = ∫a dx, which is how you handle an acceleration given in terms of position.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
a = 3t², starting from restv = t³, so v = 8 m/s at t = 2File 7 Q1
a = 6t + 4, starting at u = 2v = 3t² + 4t + 2File 7 Q2 — here C = 2, the initial velocity
v = 3t² − 2t + 1x = t³ − t² + t, so 6 m in 2 sFile 7 Q10
integrate a constant av = u + atFile 2 — this is where the first kinematic equation comes from
∫v dv = ∫4x dx from 0 to 3v²/2 = 18, v = 6 m/sFile 7 Q5 — the position route
17

Limits — squeezing the gap to nothing

Class 11, but the idea is simple
In plain words — the ten-year-old version

Suppose you want to know how fast a car was going at one exact instant, not over a whole journey. Awkward: at an instant, no time passes and no distance is covered.

So you cheat, carefully. Measure over 2 seconds. Then over 1 second. Then 0.1. Then 0.01. The answers stop jumping around and settle on one number. That settled number is the answer, and the process is called taking a limit.

On a graph you are watching a chord between two points slowly turn into a tangent at one point.

You will not have to compute limits from scratch in NEET — you use the power rule instead. But you need the idea, because it is the reason average velocity and instantaneous velocity are different things, and that distinction is examined directly.

t = 4 s shrink the gap, watch the answer: Δt = 2.0 → 3.92 Δt = 1.0 → 3.86 Δt = 0.1 → 3.84 it settles on 3.84 Squeeze the two dots together and the chord turns into the tangent. That settled value is the instantaneous velocity. It is what dx/dt means.
animated Shrink the interval and the average velocity settles down. That settled value is the instantaneous velocity.
Where you actually need it
  • The whole opening of File 1: NCERT Table 2.1 shows the answers converging on 3.84 m/s.
  • It is why v = dx/dt is written as a limit in the NCERT definition.
  • It explains why instantaneous speed always equals |instantaneous velocity| — in a vanishing interval there is no time to double back.
  • It is why a graph with a sharp corner has no defined velocity at that corner: the chord settles on two different answers depending on which side you come from.
  • Acceleration is the same idea one rung up: a = lim(Δv/Δt).

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
Δt = 2 s about t = 4 on x = 0.08t³3.92 m/sFile 1 — the first, roughest estimate
Δt = 1 s3.86 m/sFile 1 — closer
Δt = 0.1 s3.84 m/sFile 1 — settled
power rule: v = 0.24t² at t = 43.84 m/s exactlyFile 1 — the algebra agrees with the squeezing
chord slope vs tangent slopeaverage vs instantaneousFile 1 and File 6 — the distinction the whole chapter rests on
18

Exponentials and logarithms

Class 11
In plain words — the ten-year-old version

Take a number and halve it. Halve it again. And again: 20, 10, 5, 2.5, 1.25… You get very small very fast, but you never actually reach zero. That shape is called an exponential decay, written e−kt.

It shows up whenever something shrinks at a rate that depends on how big it currently is. A body slowed by a = −kv does exactly that: the faster it goes, the harder it is braked, so it eases off more and more gently.

A logarithm is the undo button. ln answers ‘what power of e gives me this?’ You need it because integrating 1/v always produces a logarithm.

The lovely paradox: this body never quite stops, yet it travels only a finite distance — u/k. Both statements are true, and NEET plays them against each other.

1052.51.25 20 Every 1.39 s the speed halves: 20, 10, 5, 2.5, 1.25 … Halving forever never quite reaches zero — but the total distance is still only 40 m. t
animated Halving forever gets close to zero but never touches it — yet the total distance covered is still finite.
Where you actually need it
  • Resistive or drag retardation, a = −kv, in File 7.
  • The conceptual pair: ‘never stops’ (in time) but ‘travels only u/k’ (in distance).
  • The same maths reappears later in radioactive decay and in capacitor discharge — learning the shape now pays out twice more.
  • Distinguishing ue−kt (function of time) from ue−kx (function of position). Both are correct for the same motion, and both appear in the options.
  • Recognising that ∫dv/v = ln v is what forces the exponential to appear at all.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
a = −kv, u = 20, k = 0.5v = 20 e−0.5tFile 7 Q7
ln(v/u) = −ktthe log form of the same thingFile 7 Q7 — the line you actually write down
halving time = ln2 ÷ k1.39 s when k = 0.5File 7 figure — the speed halves every 1.39 s
20 e−0.5×8= 0.37 m/s after 8 sFile 7 — small, but not zero
total distance = u/k= 40 mFile 7 Q6 — finite, even though it never stops
19

Rounding, decimals and how many figures to keep

Class 6–7, plus judgement
In plain words — the ten-year-old version

Rounding to two decimal places means: look at the third decimal. If it is 5 or more, push the second one up. If less, leave it. So 3.0625 becomes 3.06.

Two pieces of judgement matter more than the rule itself.

Round only at the end. If you round 3.0625 to 3.06 and then divide 35 by it, you get a slightly different answer. Keep the full number in your calculator and round the final result.

Sometimes you must round up regardless. If the answer is ‘the bullet is stopped after 10.26 planks’, there is no such thing as 0.26 of a plank — you need 11. Ask what the number is counting before you round it.

And g: use 10 when the question’s numbers are round, 9.8 when it says so or when the options are close together.

3.05 3.06 3.07 3.0625 is here It is nearer to 3.06 than to 3.07, so to two decimals it rounds DOWN to 3.06. But not always! If 10.26 planks would stop the bullet, you need 11. Sometimes you must round UP.
animated Round at the very end, not in the middle. And ask what the number counts — planks and people always round up.
Where you actually need it
  • NCERT Exercise 2.5 gives 3.0625 and 11.4285…, quoted as 3.06 and 11.4.
  • Choosing between g = 9.8 and g = 10 — look at the options before deciding.
  • Surd answers such as 2√10 may be exact in the options, so do not decimalise unless you must.
  • Counting problems (planks, bounces, complete laps) always round up to the next whole one.
  • If two options differ only in the second decimal place, that is a sign the question wants g = 9.8, not 10.

Five examples

The sumWhat it comes toWhere it turns up in Chapter 2
3.0625 to 2 decimal places3.06File 2 Q1 — the retardation of the braking car
35 ÷ 3.0625 = 11.4285…11.4 sFile 2 Q1 — round only at this final step
400 ÷ 39 = 10.26 planks11 planksFile 2 — round UP; a fraction of a plank stops nothing
√40 = 6.3245…leave it as 2√10 if that is an optionFile 7 Q14 — exact beats decimal
g = 9.8 or g = 10?10 for round numbers, 9.8 if the options are closeFile 3 — check the options before you commit
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.