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.
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.
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).
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| +20 m/s and −20 m/s | same speed, opposite ways | File 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 | unchanged | File 3 — gravity points down the whole flight, whichever way the ball moves |
| displacement −4 m distance +4 m | over t = 3 s to 5 s | File 6 Q7 — where the two averages come apart |
| |−9| | = 9 | File 7 Q3 — the speed is 9 m/s, the velocity is −9 m/s |
| vAB = +12, so vBA | = −12 | File 5 Q14 — swapping the two bodies flips the sign only |
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 sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 1000 ÷ 3600 | = 5/18 | the reason the factor exists at all — worth deriving once |
| 18 × 5/18 | = 5 m/s | the easiest one to check the rule with |
| 126 × 5/18 | = 35 m/s | File 2 Q1 — NCERT Exercise 2.5, the braking car |
| 54 × 5/18 | = 15 m/s | File 5 Q1 — NCERT Exercise 2.14, cars B and C |
| 192 × 5/18 | = 53.3 m/s | File 5 Q16 — the thief’s speeding car |
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 v², 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| distances after 1, 2, 3, 4 s from rest | 1 : 4 : 9 : 16 | File 2 — because s ∝ t² |
| distances in the 1st, 2nd, 3rd, 4th second | 1 : 3 : 5 : 7 | File 2 — Galileo’s law of odd numbers |
| speed doubles, stopping distance | × 4 | File 2 — NCERT braking data 10, 20, 34, 50 m |
| speed doubles, stopping time | × 2 | File 2 — the contrast with the line above is the whole question |
| path 12 m, displacement 6 m | ratio 2 : 1 | File 6 Q10 — the time cancels, so never compute the averages |
5² 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 35² | = 1225 | File 2 Q1 — the braking car, before dividing by 400 |
| 5² − 4² | = 25 − 16 = 9 m | File 2 Q2 — why the distance in the 5th second is 9 m |
| √(2 × 2 ÷ 12) | = 0.577 s | File 5 Q10 — the coin dropped in a lift |
| √40 | = 2√10 = 6.32 m/s | File 7 Q14 — speed from the area under an a–x graph |
| √250 | = 15.81 m/s | File 2 — a v² = u² + 2as answer that stays untidy |
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 u² 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| v² = u² + 2as, make a the subject | a = (v² − u²)/2s | File 2 Q1 — finding the retardation of the braking car |
| s = ut + ½at², make a the subject | a = 2(s − ut)/t² | File 5 Q1 — the minimum acceleration in NCERT Exercise 2.14 |
| v = u + at, make t the subject | t = (v − u)/a | File 2 — every ‘how long to stop?’ question |
| H = u²/2g, make u the subject | u = √(2gH) | File 3 — the launch speed needed for a given height |
| 1/t = 1/t₁ + 1/t₂, make t the subject | t = t₁t₂/(t₁ + t₂) | File 5 Q6 — the escalator; you must flip at the end |
An equation with one t in it and no t² 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 0 = 35 − 3.0625 t | t = 11.4 s | File 2 Q1 — how long the car takes to stop |
| 1000 = 200 + 800 a | a = 1 m s⁻² | File 5 Q1 — NCERT Exercise 2.14 |
| 20t − 5t² = 20(t−2) − 5(t−2)² | t = 3 s | File 5 Q12 — the t² terms cancel, leaving a linear equation |
| 2 − 0.5 t = 0 | t = 4 s | File 7 Q11 — where the velocity is greatest |
| 6 − 2 t = 0 | t = 3 s | File 6 Q7 — where the particle turns round |
An equation with a t² 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 5t² − 30t + 40 = 0 | t = 2 s and t = 4 s | File 3 — the ball is at 40 m twice |
| t² − 10t − 75 = 0 | t = 15 s (reject −5) | File 5 Q8 — the police jeep catching the thief |
| 5t² − 20t − 25 = 0 | t = 5 s (reject −1) | File 3 — NCERT Example 2.3, the ball from the rooftop |
| t(2t − 10) = 0 | t = 0 or t = 5 s | File 5 Q13 — t = 0 is the start, so the meeting is at 5 s |
| t(10 − t²/3) = 0 | t = √30 = 5.48 s | File 7 Q15 — returning to the origin under a = −2t |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 20 + 15t = 140 + 3t | t = 10 s, x = 170 m | File 5 Q11 — P overtakes Q |
| t² = 75 + 10t | t = 15 s | File 5 Q8 — jeep from rest vs thief at steady speed |
| 20t − 5t² = 20(t−2) − 5(t−2)² | t = 3 s, height 15 m | File 5 Q12 — two stones thrown 2 s apart |
| ½(4)t² = 10t | t = 5 s | File 5 Q13 — accelerating body catches a steady one |
| 50t + 50t = 100 km | t = 1 h | File 5 Q9 — two cars approaching, which fixes how long the bird flies |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| read x at t = 4 s on x = 0.08t³ | x = 5.12 m | File 1 — NCERT Table 2.1 |
| where an x–t line meets the up-axis | that value is x₀ | File 4 — the starting position, often not zero |
| where a v–t line crosses the time axis | v = 0, the turning point | File 6 Q7 — split the journey here |
| where two x–t lines cross | same position → they meet | File 4 & File 5 Q11 |
| axis says ‘x’ or axis says ‘v’ | completely different meaning | File 4 — the most common misread in the chapter |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| (33 − 5) ÷ (3 − 1) | = 14 m/s | File 6 Q6 — chord slope, so this is the average velocity |
| tangent slope at t = 4 s on x = 0.08t³ | = 3.84 m/s | File 1 — tangent slope, so this is the instantaneous velocity |
| slope of a v–t line | = acceleration | File 4 — a straight v–t line means constant acceleration |
| slope of an x–t line = 0 | the body is at rest | File 4 — a flat line is not ‘no graph’, it is standing still |
| 15 − 3 | = 12 m/s relative | File 5 Q11 — relative velocity is the gap between two slopes |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 6 m/s held for 4 s | rectangle = 24 m | File 2 — the v₀t part of the second equation |
| ½ × 4 × 12 | triangle = 24 m | File 2 — the ½at² part, the bonus the acceleration buys |
| ½ × (6 + 18) × 4 | trapezium = 48 m | File 4 Q17 — both halves in one step |
| areas +9 m and −4 m | displacement 5 m, distance 13 m | File 6 Q7 — the reversal splits the two answers |
| area under a–t of 2 m/s² for 2 s | Δv = 4 m/s | File 4 — a–t area gives velocity, not distance |
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 t², its graph is a straight line. That is what steady speed looks like: equal steps in equal times.
If it has a t² in it, the graph is a parabola — a curve that gets steeper and steeper. That is what speeding up looks like. If the t² has a minus in front, the parabola opens downwards, like the path of a thrown ball.
So: x = v₀t + ½at² has a t² in it, which is exactly why every constant-acceleration x–t graph is a curve, never a line.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| x = 20 + 15t | straight line | File 5 Q11 — no t², so steady speed |
| x = v₀t + ½at² | parabola | File 2 — the t² is what makes it bend |
| v = v₀ + at | straight line | File 4 — velocity climbs steadily under constant acceleration |
| a = −9.8 in free fall | flat horizontal line | File 3 — the a–t graph is the simplest of the three |
| y = 25 + 20t − 5t² | downward parabola | File 3 — NCERT Example 2.3, the ball from the rooftop |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| distances after 1, 2, 3, 4 s | 1 : 4 : 9 : 16 | File 2 — the totals, because s ∝ t² |
| gaps between those totals | 1 : 3 : 5 : 7 | File 2 — Galileo’s odd numbers |
| S₅ with u = 0, a = 2 | = (2/2)(2×5 − 1) = 9 m | File 2 Q2 — the 5th second, not the first five seconds |
| check it: 5² − 4² | = 25 − 16 = 9 m | File 2 Q2 — the subtraction check, always worth doing |
| distance in the last second before stopping | = a/2, whatever u was | File 2 — comes from running the motion backwards |
‘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 mean — 2v₁v₂/(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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| equal times at 40 and 60 | (40+60)/2 = 50 km/h | File 6 Q3 — arithmetic mean |
| equal distances at 40 and 60 | 2(40)(60)/100 = 48 km/h | File 6 Q2 — harmonic mean, always the smaller |
| equal distances at 20, 30, 60 | 3 ÷ (1/20+1/30+1/60) = 30 | File 6 Q4 — harmonic mean of three |
| mid-TIME velocity, u = 4, v = 16 | (4+16)/2 = 10 m/s | File 6 Q11 — arithmetic |
| mid-POINT velocity, u = 4, v = 16 | √[(16+256)/2] = 11.66 m/s | File 2 — root-mean-square, always the larger |
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 t³ becomes 3t². And t² becomes 2t. And t (which is secretly t¹) 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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| x = 0.08t³ | v = 0.24t² | File 1 — NCERT Table 2.1; at t = 4 this gives 3.84 m/s |
| x = 8 + 12t − t³ | v = 12 − 3t², a = −6t | File 1 — an AIPMT item; the 8 disappears |
| x = 3t² + 2t | v = 6t + 2 | File 6 Q6 — used to check the chord answer of 14 m/s |
| x = t³ − 6t² + 3t + 4 | v = 3t² − 12t + 3 a = 6t − 12 | File 7 Q3 — a = 0 at t = 2 s |
| differentiate the number 4 | = 0 | File 7 — a constant starting position never affects the velocity |
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 t³.
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| a = 3t², starting from rest | v = t³, so v = 8 m/s at t = 2 | File 7 Q1 |
| a = 6t + 4, starting at u = 2 | v = 3t² + 4t + 2 | File 7 Q2 — here C = 2, the initial velocity |
| v = 3t² − 2t + 1 | x = t³ − t² + t, so 6 m in 2 s | File 7 Q10 |
| integrate a constant a | v = u + at | File 2 — this is where the first kinematic equation comes from |
| ∫v dv = ∫4x dx from 0 to 3 | v²/2 = 18, v = 6 m/s | File 7 Q5 — the position route |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| Δt = 2 s about t = 4 on x = 0.08t³ | 3.92 m/s | File 1 — the first, roughest estimate |
| Δt = 1 s | 3.86 m/s | File 1 — closer |
| Δt = 0.1 s | 3.84 m/s | File 1 — settled |
| power rule: v = 0.24t² at t = 4 | 3.84 m/s exactly | File 1 — the algebra agrees with the squeezing |
| chord slope vs tangent slope | average vs instantaneous | File 1 and File 6 — the distinction the whole chapter rests on |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| a = −kv, u = 20, k = 0.5 | v = 20 e−0.5t | File 7 Q7 |
| ln(v/u) = −kt | the log form of the same thing | File 7 Q7 — the line you actually write down |
| halving time = ln2 ÷ k | 1.39 s when k = 0.5 | File 7 figure — the speed halves every 1.39 s |
| 20 e−0.5×8 | = 0.37 m/s after 8 s | File 7 — small, but not zero |
| total distance = u/k | = 40 m | File 7 Q6 — finite, even though it never stops |
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.
| The sum | What it comes to | Where it turns up in Chapter 2 |
|---|---|---|
| 3.0625 to 2 decimal places | 3.06 | File 2 Q1 — the retardation of the braking car |
| 35 ÷ 3.0625 = 11.4285… | 11.4 s | File 2 Q1 — round only at this final step |
| 400 ÷ 39 = 10.26 planks | 11 planks | File 2 — round UP; a fraction of a plank stops nothing |
| √40 = 6.3245… | leave it as 2√10 if that is an option | File 7 Q14 — exact beats decimal |
| g = 9.8 or g = 10? | 10 for round numbers, 9.8 if the options are close | File 3 — check the options before you commit |