Chapter 2 of 14

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Haryana (BSEH)Class 11 Physics★ 5-7 marks · CBSE Class 11 Physics Chapter 2; kinematics is foundational for JEE and NEET

Motion in a Straight Line

Drop a ruler through a friend's fingers, measure how far it fell before they caught it, and you have measured their reaction time. This chapter builds the tools for exactly that — velocity at a single instant, found by shrinking a time window until the number stops moving; Galileo settling whether acceleration is measured against distance or against time; and three equations derived twice over, once from the area under a graph and once by integration. It never asks what causes motion. That is the next chapter. CBSE Class 11 Physics Chapter 2.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

Podcast

Listen as a conversation

A teacher & a student talk through the chapter — in your language.

Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

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VocabularyBuild your vocabularyLexiGenome mines the key words and schedules them with spaced repetition.Mine words →

By the end of this chapter you'll be able to…

1Define a frame of reference, and say why motion is meaningless without one
2Find instantaneous velocity as a limit, and read it as the slope of a position-time graph
3Tell average speed from the magnitude of average velocity
4Use the three kinematic equations, and know when they stop being valid
5Derive them both ways — area under a graph, and by calculus
6Apply them to free fall, keeping the sign convention straight

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Why this chapter matters

Describing motion turns out to need one idea the algebra hides: velocity at a single instant. This chapter gets there by shrinking a time window until the number stops moving, then builds the three equations everything else in mechanics rests on. It never asks what causes motion — that is Chapter 4.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Units and Measurements

Significant figures and dimensional checking, used on every numerical answer here

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Class 9 Science: Motion

Distance, displacement, speed, velocity and the basic equations of motion, treated qualitatively

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Class 11 Mathematics: Limits and Derivatives

The limit as delta-t tends to zero, and differentiation, are used directly to define velocity and acceleration

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Class 11 Mathematics: Integrals

Needed for Example 2.2, which re-derives the kinematic equations by integration

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Class 10 Mathematics: Quadratic Equations

Example 2.3's second method solves a quadratic in t to find total flight time in one step

Motion in a Straight Line

1. Before you start: two gaps between CBSE and the textbook

Check these first, because your notes almost certainly do not mention them.

Topic	NCERT chapter	CBSE 2026-27	What it means for you
Frame of reference	Never defined. The word "frame" does not appear once.	Listed as examinable	You can be asked about it, and the chapter will not help. See section 8.
Relative velocity	Listed as "2.5" in the contents page, but there is no such section in the body	Not listed for this chapter	Not examinable here — but Exercise 2.14 still needs it. See section 7.

Everything else in the chapter maps cleanly:

Textbook section	Topic
2.1	Introduction, and the point-object approximation
2.2	Instantaneous velocity and speed
2.3	Acceleration
2.4	Kinematic equations for uniformly accelerated motion

One line from the introduction is worth holding onto, because students forget it under pressure: kinematics describes motion without ever asking what causes it. Nothing in this chapter explains why anything accelerates. That is Chapter 4's job.

2. Four things the textbook keeps warning you about

The chapter closes with six Points to Ponder. Four of the six — points 1, 2, 3 and 5 — are about signs and direction. That is what the book itself keeps returning to, and none of it is algebra.

Mistake 1: reading a minus sign as "slowing down"

This is the big one, and the textbook devotes three separate Points to Ponder to it.

A negative acceleration does not mean an object is slowing down. The sign only records which direction you chose as positive.

Take upward as positive. Gravity is then $a = - g$ , always, for the whole flight:

Stage of a thrown ball	Velocity	Acceleration	Speed is…
Going up	positive	$- g$	decreasing
At the top	zero	$- g$	momentarily zero
Coming down	negative	$- g$	increasing

Same negative acceleration throughout. It slows the ball on the way up and speeds it up on the way down.

The rule that actually works: compare the directions of velocity and acceleration.

Same direction → speeding up
Opposite directions → slowing down

That statement does not depend on which way you called positive, which is exactly why it is the one to memorise.

So state your origin and positive direction before anything else. Every sign in the working depends on that choice, and the book puts this first for the same reason.

Mistake 2: thinking zero velocity means zero acceleration

At the highest point of a throw the ball is momentarily at rest. Students conclude nothing is acting on it.

Gravity never switches off. If acceleration really became zero at the top, the ball would stay there.

Velocity passes through zero. Acceleration does not. On an acceleration-time graph for the whole flight you draw one flat line at $- g$ , with no break anywhere in it.

Mistake 3: treating average speed as the size of average velocity

Walk to a shop and back. Your displacement is zero, so your average velocity is zero. Your average speed is not.

$average speed \geq ∣ average velocity ∣$

They are equal only when the motion never reverses direction.

Exercise 2.10 makes the point sharply — you would not want to tell a man who walked to the market and back that his average speed was zero.

But instantaneously the problem vanishes. At a single instant there is no path to average over, so instantaneous speed always equals the magnitude of instantaneous velocity. That is what Exercise 2.11 is asking you to explain.

Mistake 4: using the kinematic equations when acceleration is not constant

$v = v_{0} + a t$ and its two companions were derived by treating $a$ as a fixed constant during integration. Apply them to varying acceleration and they are simply false.

When acceleration varies, go back to the definition and integrate:

$a = \frac{d v}{d t} ⟹ v = \int a d t$

That is precisely why the textbook bothers to derive the equations a second time using calculus.

3. Velocity at an instant

Average velocity over an interval hides everything that happened inside it. To get velocity at one instant, shrink the interval:

$v = lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t}$

The chapter makes that limit concrete rather than abstract. For $x = 0.08 t^{3}$ , it computes the average velocity over windows centred on $t = 4$ s, shrinking each time — 2.0 s, then 1.0, 0.5, 0.1, and finally 0.01 s. The value marches steadily onto 3.84 m/s and stops moving.

That number is $d x / d t$ at $t = 4$ s. You can watch the limit converge instead of taking it on faith.

Graphically: instantaneous velocity is the slope of the tangent to the position-time graph. Acceleration is the slope of the tangent to the velocity-time graph.

Worked: textbook example 2.1

Given $x = a + b t^{2}$ with $a = 8.5$ m and $b = 2.5$ m/s².

Differentiate: $v = \frac{d x}{d t} = 2 b t = 5.0 t$ m/s. So $v = 0$ at $t = 0$ , and $v = 10$ m/s at $t = 2.0$ s.

Average velocity from $t = 2.0$ s to $t = 4.0$ s is $\frac{x ( 4 ) - x ( 2 )}{2} = \frac{12 b}{2} = 15$ m/s.

Notice it matches neither endpoint's instantaneous value. That mismatch is normal whenever velocity is changing — and it is the whole reason the two words exist.

4. Why acceleration is defined against time

This was genuinely unsettled in Galileo's day, and it is worth knowing why it went the way it did.

The open question was whether to define acceleration as the rate of change of velocity with distance or with time. Galileo's work on falling bodies and inclined planes settled it: with time, the rate is constant for all objects in free fall. With distance, it is not — it decreases as the fall continues.

So the definition was chosen because it produces a quantity that stays fixed in the most important case:

$a = lim_{Δ t \to 0} \frac{Δ v}{Δ t} = \frac{d v}{d t}$

5. The kinematic equations, and where they come from

The chapter derives them twice. Knowing both routes matters, because either can be asked.

Route 1 — area under the velocity-time graph. For constant acceleration the graph is a straight line. The area under it is a rectangle plus a triangle:

$x = v_{0} t + \frac{1}{2} (v - v_{0}) t = v_{0} t + \frac{1}{2} a t^{2}$

Route 2 — calculus. Integrate the definitions directly:

$\int_{v_{0}}^{v} d v = \int_{0}^{t} a d t ⟹ v = v_{0} + a t$

$\int_{x_{0}}^{x} d x = \int_{0}^{t} (v_{0} + a t) d t ⟹ x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$

Using the chain rule to swap variables, $a = v \frac{d v}{d x}$ , gives the third:

$v^{2} = v_{0}^{2} + 2 a (x - x_{0})$

The calculus route survives non-constant acceleration. The area route does not. That is the advantage the textbook points out, and the reason it does the work twice.

Together the three connect five quantities — $v_{0}$ , $v$ , $a$ , $t$ , $x$ . Know any three and the rest follow.

Why the area is a displacement at all: the vertical axis is m/s, the horizontal is s. Multiply them and the seconds cancel, leaving metres. An area under a graph is only ever a physical quantity because of what the axes multiply out to.

6. Free fall, and two examples worth copying

Free fall is not a new topic. It is the same three equations with $v_{0} = 0$ and $a = - g$ , taking upward as positive:

$v = - g t y = - \frac{1}{2} g t^{2} v^{2} = - 2 g y$

The two-method problem (example 2.3)

A ball is thrown up at 20 m/s from a 25 m building. How long before it hits the ground?

Split-the-path method: find the time up (2 s), then the time falling from the 45 m peak (3 s). Total 5 s.

Single-equation method: put in the start and end states and solve one quadratic.

$0 = 25 + 20 t - 5 t^{2} ⟹ t = 5 s$

The textbook says outright that the second is better — the equations only need the start and end states and do not care what happened in between. Fewer steps, fewer sign errors.

Galileo's law of odd numbers (example 2.5)

A body dropped from rest covers distances in successive equal time intervals in the ratio 1 : 3 : 5 : 7.

Position after $n$ intervals goes as $n^{2}$ , so the distance during the $n$ -th interval alone is $n^{2} - (n - 1)^{2} = 2 n - 1$ . Run $n = 1, 2, 3, 4$ and you get 1, 3, 5, 7.

It falls straight out of distance growing as $t^{2}$ — nothing more exotic.

Stopping distance (example 2.6)

$d_{s} = \frac{v _{0}^{2}}{2 a}$

Stopping distance goes as the square of speed. Double your speed and you need four times the distance to stop.

The chapter backs this with real measured data for one car: 10, 20, 34 and 50 m at 11, 15, 20 and 25 m/s. That square law is the physics behind school-zone speed limits.

7. What Exercise 2.14 needs

The relative velocity section was removed from the body, but the exercise survived. Here is the minimum.

Definition: the velocity of A as measured by an observer moving with B is $v_{A B} = v_{A} - v_{B}$ , taken as signed quantities along one axis.

Situation	Relative speed
Same direction	speeds subtract
Opposite directions	speeds add

Exercise 2.14 worked. A police van moving at 30 km/h fires a bullet with muzzle speed 150 m/s at a car fleeing at 192 km/h in the same direction.

The bullet's ground speed is $150 + 8.33 = 158.33$ m/s. The car's is 53.33 m/s. What damages the car is the bullet's speed relative to the car:

$158.33 - 53.33 \approx 105 m/s$

Not the muzzle speed, and not the ground speed. Fix a positive direction before you subtract anything.

8. Frame of reference — examinable, but not in the chapter

CBSE lists it under this chapter. The NCERT chapter never defines it, so here it is.

A frame of reference is the coordinate system, with an origin and a chosen positive direction, that an observer uses to measure position and time.

The chapter uses one throughout without naming it — it tells you to specify position "with reference to a conveniently chosen origin", and to take one direction as positive.

Why it matters: motion has no meaning without stating the frame. A passenger sitting on a moving train is at rest relative to the train and moving at 80 km/h relative to the platform. Both are correct. The question is only ever incomplete if the frame is missing.

This is also the idea underneath relative velocity in section 7: changing frames changes the measured velocity, and $v_{A B} = v_{A} - v_{B}$ is the rule for converting between them.

9. Summary

Kinematics describes motion without asking what causes it — that is Chapter 4's job.
Instantaneous velocity $v = d x / d t$ is the slope of the tangent to the position-time graph; instantaneous acceleration $a = d v / d t$ is the slope of the tangent to the velocity-time graph.
Average speed is always $\geq$ the magnitude of average velocity; the two coincide only instantaneously, or when the motion never reverses direction.
A negative acceleration does not mean "slowing down" — compare the directions of velocity and acceleration, not the sign of acceleration alone.
Zero velocity does not imply zero acceleration: at the top of a throw, gravity never switches off.
The three kinematic equations — $v = v_{0} + a t$ , $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ , $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ — hold only while acceleration is constant; otherwise, integrate $a = d v / d t$ directly.
They can be derived two ways: geometrically, from the area under a v-t graph, or by calculus. Only the calculus route survives non-constant acceleration.
Free fall is not new physics — it is the same three equations with $v_{0} = 0$ and $a = - g$ .
Stopping distance grows as the square of speed: $d_{s} = v_{0}^{2} /2 a$ .
Galileo settled that acceleration is defined against time, not distance, because only the time-based rate stays constant in free fall.
Relative velocity, $v_{A B} = v_{A} - v_{B}$ , is not taught in the 2026-27 body but is still needed for Exercise 2.14.
Frame of reference is examinable under CBSE 2026-27 even though the NCERT chapter never defines the term.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Point-object approximation

valid while the object's size is much smaller than the distance it moves

Stated before anything else in the chapter; it is an assumption, not a fact

Average velocity

v_avg = (displacement) / (time interval)

Hides everything that happened inside the interval

Instantaneous velocity

v = lim (delta-x / delta-t) as delta-t tends to 0 = dx/dt

The slope of the TANGENT to the x-t graph at that instant

Speed vs velocity, averaged

average speed >= |average velocity|

Walk to a shop and back: average velocity is zero, average speed is not

Speed vs velocity, instantaneous

instantaneous speed = |instantaneous velocity|, always

At one instant there is no path to average over, so they coincide

Average acceleration

a_avg = (v2 - v1) / (t2 - t1) = delta-v / delta-t

Slope of the straight line joining two points on a v-t graph

Instantaneous acceleration

a = lim (delta-v / delta-t) as delta-t tends to 0 = dv/dt

Slope of the tangent to the v-t curve; SI unit m/s^2

Why acceleration uses time, not distance

dv/dt is constant in free fall; dv/dx is not

Galileo's result, and the reason for the definition

Area under a v-t curve

area between t1 and t2 = displacement over that interval

Works because (m/s) x (s) = m; the seconds cancel

First kinematic equation

v = v0 + at

From integrating a = dv/dt with a constant

Second kinematic equation

x = x0 + v0*t + (1/2)a*t^2

From integrating v = dx/dt; the area under the v-t line

Third kinematic equation

v^2 = v0^2 + 2a(x - x0)

From v dv = a dx, or by eliminating t between the first two

Average velocity, constant a only

v_avg = (v0 + v)/2

The arithmetic mean works ONLY when acceleration is constant

Free fall (upward positive, from rest)

v = -g*t; y = -(1/2)g*t^2; v^2 = -2g*y

Not a new topic: the same three equations with v0 = 0 and a = -g

Galileo's law of odd numbers

distances in successive equal times are as 1 : 3 : 5 : 7 ...

The n-th interval covers (2n-1) units, because distance grows as t^2

Stopping distance

d_s = v0^2 / (2a)

Doubling the speed QUADRUPLES the stopping distance

Reaction time from a dropped ruler

t_r = sqrt(2d/g)

d = 21.0 cm gives about 0.21 s

Distance in the nth second

s_n = u + a(n - 1/2)

A shortcut, derived from the second kinematic equation

Relative velocity

v_AB = v_A - v_B (signed, along one axis)

Same direction: speeds subtract. Opposite: speeds add

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Assuming frame of reference is not examinable because the chapter never mentions it

✓ CBSE 2026-27 lists 'Frame of reference' under this chapter, but the word 'frame' does not appear once in the NCERT text. Learn it separately: a frame of reference is the coordinate system, with an origin and a chosen positive direction, that an observer measures position and time against. A passenger is at rest relative to the train and moving relative to the platform — both are correct, and a question is incomplete without naming the frame.

WATCH OUT

✗ Confusing path length (distance) with displacement

✓ Displacement is the change in position and can be zero for a round trip while the distance walked is large. Exercise 2.9 asks you to prove the general inequality between them and state exactly when equality holds — when the motion never reverses direction.

WATCH OUT

✗ Assuming average speed always equals the magnitude of average velocity

✓ It is greater than OR equal to it. Exercise 2.10's note makes the point sharply: you would not want to tell a tired man who walked to the market and back that his average speed was zero. Only for instantaneous values do speed and |velocity| always coincide.

WATCH OUT

✗ Using v = v0 + at when acceleration is not constant

✓ All three kinematic equations were derived by treating a as a fixed constant during integration, so they are simply false otherwise. Go back to a = dv/dt and integrate the actual a(t), which is exactly why Example 2.2 bothers with the calculus route.

WATCH OUT

✗ Reading 'negative acceleration' as 'slowing down'

✓ The sign only records which direction you chose as positive. Take upward as positive and gravity is always negative — yet a falling object speeds up under it while a rising object slows down under that very same negative acceleration.

WATCH OUT

✗ Assuming zero velocity means zero acceleration

✓ A ball thrown straight up has v = 0 for one instant at the top, but gravity never switches off during the flight. If it did, the ball would simply hover there. This is Points to Ponder #4 and a standard exam trap.

WATCH OUT

✗ Treating the area under a v-t graph as distance

✓ It is displacement. Where velocity is negative the area counts negative, exactly as a signed integral should. Total path length is a separate, always-positive calculation that never allows cancellation.

WATCH OUT

✗ Forgetting to fix the origin and positive direction before substituting numbers

✓ Points to Ponder #1: every sign that follows — of displacement, velocity and acceleration — depends on that choice. State it in the first line of the answer, or the signs in your working mean nothing.

WATCH OUT

✗ Splitting a vertical-motion problem into up and down phases when you do not need to

✓ Example 2.3 solves the same question both ways and states outright that the single-equation method is better, because the kinematic equations only need the start and end states and do not care what path was taken in between.

WATCH OUT

✗ Using the arithmetic mean (v0 + v)/2 for average velocity when acceleration varies

✓ That shortcut is derived from the straight-line v-t graph of constant acceleration. With a curved v-t graph it is simply wrong; you must integrate instead.

WATCH OUT

✗ Reading a sharp kink on a graph as physically real

✓ The chapter says these are idealisations only: in any realistic situation the functions are differentiable everywhere and the graphs are smooth, because velocity and acceleration cannot change abruptly at an instant.

WATCH OUT

✗ Quoting the bullet's muzzle speed as the answer to Exercise 2.14

✓ What damages the thief's car is the bullet's speed RELATIVE TO THAT CAR, not its muzzle speed or its ground speed. Convert everything to one unit, fix a positive direction, then subtract.

WATCH OUT

✗ Adding speeds for two objects travelling in the same direction

✓ Relative velocity is v_A - v_B as signed quantities. Same direction means the speeds subtract (a car doing 80 seen from one doing 60 appears to move at 20); only opposite directions make them add.

WATCH OUT

✗ Believing the chapter still teaches relative velocity in full

✓ Section 2.5 survives only as a contents-page label and one line in the introduction — no formula, derivation or worked example appears in the 2026-27 body. Exercise 2.14 still needs the concept, so learn it from the chapter notes.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Position-time graphs, kinematic equations, free fall, relative motion, and speed-vs-velocity reasoning (Q2.1-2.18)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Motion in a Straight Line?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~7 marks in Haryana (BSEH) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Frame of reference: an origin, a positive direction, a clock. CBSE examines it; the chapter never defines it.
•Kinematics describes motion and never asks what causes it.
•Instantaneous velocity v = dx/dt — the slope of the tangent to the position-time graph.
•Instantaneous acceleration a = dv/dt — the slope of the tangent to the velocity-time graph.
•Average speed is greater than or equal to the magnitude of average velocity; equal only if the motion never reverses.
•Instantaneous speed always equals the magnitude of instantaneous velocity.
•Acceleration is defined against time, not distance — Galileo's result for free fall.
•Area under a velocity-time curve is displacement, because m/s times s leaves metres.
•v = v0 + at; x = x0 + v0t + at^2/2; v^2 = v0^2 + 2a(x - x0).
•Those three hold only while acceleration is constant. Otherwise integrate a = dv/dt.
•A negative acceleration does not mean slowing down — compare the directions of v and a.
•Same direction means speeding up; opposite directions means slowing down.
•Zero velocity does not imply zero acceleration — a ball at the top of a throw still has g.
•State the origin and positive direction before assigning any sign.
•Free fall, upward positive, from rest: v = -gt, y = -gt^2/2, v^2 = -2gy.
•Galileo's odd numbers: distances in successive equal times go 1 : 3 : 5 : 7.
•Stopping distance = v0^2 / 2a, so it grows with the square of speed.
•Relative velocity v_AB = v_A - v_B. Not examinable here, but Exercise 2.14 needs it.

Haryana (BSEH) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-7 marks across the chapter; kinematics is examined every year

Question type	Marks each	Typical count	What it tests
Instantaneous velocity, acceleration and graphs	1-2	1-2	Reading slope vs area on x-t, v-t and a-t graphs; zero velocity with non-zero acceleration; distinguishing average from instantaneous velocity
Kinematic equations and free fall	2-3	1-2	Applying v=u+at, s=ut+½at², v²=u²+2as to free fall and braking problems; Galileo's law of odd numbers; stopping distance's square-law dependence on speed
Relative velocity and frame of reference	3-5	1	Computing relative velocity of two moving bodies along one axis; explaining frame of reference and why motion needs one stated; full vertical-motion or graph-derivation problems

Prep strategy

Learn the chapter as three blocks: velocity at an instant (2.2), acceleration and graphs (2.3), and the equations plus their applications (2.4)
Be able to derive the kinematic equations BOTH ways — geometrically from the v-t area and by integration — since either can be asked
Work all seven of the chapter's examples before touching the exercises; four of them (2.3 to 2.7) are the templates for the standard numericals
Drill the Points to Ponder list, especially #3 and #4 — the sign of acceleration, and zero velocity with non-zero acceleration, are the two most examined traps
For every graph question, practise saying out loud whether the answer comes from a slope or from an area before you compute anything
Learn relative velocity from the chapter notes, since the textbook body no longer covers it but Exercise 2.14 still requires it

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Speed limits outside schools

Stopping distance grows as the square of speed, so a modest cut in the limit produces a disproportionately large cut in the distance a car needs to stop.

Your own reaction time

Example 2.7 is a real experiment: have someone drop a ruler through your fingers, measure how far it fell before you caught it, and t = sqrt(2d/g) gives your reaction time — about 0.2 s for most people.

Anti-lock braking systems

ABS exists because braking performance depends on maintaining a predictable deceleration; the stopping-distance formula assumes exactly that constant a.

Skydiving and terminal velocity

Free fall as taught here neglects air resistance; comparing the idealised prediction against what skydivers actually experience is how terminal velocity is introduced.

Measuring g in the laboratory

Timing a body falling through a known height and inverting y = -(1/2)gt^2 is the standard school measurement of the acceleration due to gravity.

Sprint and athletics analysis

Coaches split a race into acceleration and constant-velocity phases, exactly the multi-stage v-t graph this chapter teaches you to read.

Traffic accident reconstruction

Investigators work backwards from skid-mark length using d_s = v0^2/2a to estimate how fast a vehicle was travelling before braking.

Lift and elevator design

Comfort limits are written as maximum acceleration and maximum jerk, which is why real velocity-time profiles are smoothed rather than the sharp-kinked idealisations drawn in textbooks.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State the origin and the positive direction in the first line of every answer; every sign afterwards depends on that choice

List v0, v, a, t and x with signs before touching an equation, then pick the equation that omits the quantity you neither know nor want

Prefer the single-equation method over splitting a flight into up and down phases — Example 2.3 shows it is fewer steps and fewer sign errors

For graph questions, say explicitly whether you are reading a slope or an area, and state which quantity that gives

Never quote average velocity as (v0 + v)/2 unless you have said acceleration is constant

When asked whether a body speeds up or slows down, compare the DIRECTIONS of velocity and acceleration, never the sign of acceleration alone

In relative-velocity problems, convert every speed to one unit first, then fix a positive direction, then subtract — do not add speeds by reflex

Quote the limitation with any kinematic-equation derivation: it holds only while acceleration is constant in magnitude and direction

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive the kinematic equations for a body whose acceleration varies linearly with time, a = kt, and show where the constant-a forms break down

STRETCH

A body falls from rest under gravity with air resistance proportional to velocity. Set up the differential equation, solve for v(t), and identify the terminal velocity

STRETCH

Prove Galileo's law of odd numbers in reverse: given that successive equal-time distances are in the ratio 1:3:5:7, show the acceleration must be constant

STRETCH

Two bodies are thrown vertically upward from the same point at different times with different speeds. Find the condition on speeds and delay for them to collide in mid-air

STRETCH

The chapter states that average speed is greater than or equal to the magnitude of average velocity. Prove this for arbitrary one-dimensional motion and characterise exactly when equality holds

STRETCH

A particle moves so that its velocity is proportional to the square root of the distance covered. Find x(t), and comment on what happens at t = 0

STRETCH

Reconstruct a vehicle's speed before braking from a skid mark of known length, then estimate how sensitive your answer is to a ten per cent error in the assumed friction deceleration

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainStone dropped from a rising balloonSign-convention problem

A balloon is rising vertically with a velocity of 8 m/s. When it is at a height of 12 m above the ground, a stone is dropped from it. Find the time taken by the stone to reach the ground. (Take g = 9.8 m/s^2.)

Stuck? Show the approach

At the instant of release the stone shares the balloon's velocity, so its initial velocity is +8 m/s (upward positive), not zero. Write the position equation measured from the ground and solve the resulting quadratic in t.

Show the full solution

$y = 12 + 8 t - \frac{1}{2} (9.8) t^{2} = 0 ⟹ 4.9 t^{2} - 8 t - 12 = 0$ $t = \frac{8 + 64 + 4 ( 4.9 ) ( 12 )}{2 ( 4.9 )} = \frac{8 + 299.2}{9.8} \approx 2.58 s$

Answer: approximately 2.58 s

The trap

Taking the stone's initial velocity as zero because it is 'dropped' is the standard error — dropped means released with no push of its own, not released from rest in the ground frame. It carries the balloon's own velocity at the instant of release.

JEE MainDistance in the last two secondsKinematic equation application

A particle starts from rest and moves with constant acceleration. Its velocity after 10 s is 20 m/s. Find the distance it travels in the last 2 seconds of this interval.

Stuck? Show the approach

Find the constant acceleration from the given velocity, then compute total distance at t = 10 s and at t = 8 s separately and subtract — do not use a per-second shortcut formula for a two-second interval.

Show the full solution

$a = v / t = 20/10 = 2 m/s^{2}$ . $x (10) = \frac{1}{2} (2) (10)^{2} = 100 m$ . $x (8) = \frac{1}{2} (2) (8)^{2} = 64 m$ . Distance in the last 2 s $= 100 - 64 = 36 m$ .

Answer: 36 m

The trap

Plugging n = 2 directly into the distance-in-the-nth-second formula s_n = u + a(n - 1/2) gives the distance in the 2nd second of motion, not the distance in the last 2 seconds of a 10-second interval — the two questions look similar but are not the same calculation.

JEE MainAverage speed over a trapezoidal v-t profileGraph-based motion

A car accelerates uniformly from rest to 20 m/s in 5 s, travels at that constant speed for the next 10 s, and then decelerates uniformly to rest in a further 5 s. Find the total distance covered and the average speed for the whole journey.

Stuck? Show the approach

Sketch the v-t graph: it is a trapezoid (a triangle, a rectangle, then a triangle). Find the area of each stage separately and add them, then divide the total distance by the total time.

Show the full solution

Stage 1 (0-5 s): area $= \frac{1}{2} (5) (20) = 50$ m. Stage 2 (5-15 s): area $= 10 \times 20 = 200$ m. Stage 3 (15-20 s): area $= \frac{1}{2} (5) (20) = 50$ m. Total distance $= 50 + 200 + 50 = 300$ m over 20 s, so average speed $= 300/20 = 15$ m/s.

Answer: 300 m; average speed 15 m/s

The trap

Averaging the three stage speeds as (accelerating-average + 20 + decelerating-average)/3 ignores that the stages last different lengths of time — average speed must be computed as total distance over total time, not as an average of interval averages.

JEE AdvancedDoes the pursuing car ever catch up?Multi-step reasoning with a trap

Car B is 500 m ahead of car A on a straight road and starts from rest with a uniform acceleration of 2 m/s^2. At the same instant car A passes the starting point moving at a constant 30 m/s in the same direction. Does A ever catch up with B? If not, find the minimum separation between them and when it occurs.

Stuck? Show the approach

Write both position functions from the same origin and instant, set up the gap g(t) = x_B(t) - x_A(t), and examine it as a quadratic in t rather than assuming a crossing exists. A quadratic with no real positive root means the gap is never zero.

Show the full solution

$x_{A} = 30 t$ , $x_{B} = 500 + t^{2}$ . Gap: $g (t) = 500 + t^{2} - 30 t$ . Setting $g (t) = 0$ : $t^{2} - 30 t + 500 = 0$ , discriminant $= 900 - 2000 = - 1100 < 0$ , so there is no real solution — A never catches B.

The gap $g (t) = t^{2} - 30 t + 500$ is an upward parabola, minimum at $t = 15$ s: $g (15) = 225 - 450 + 500 = 275$ m.

Answer: A never catches B; minimum separation is 275 m, occurring at t = 15 s

The trap

Assuming a catch-up time must exist just because B starts from rest while A is already moving is the trap — B's acceleration means its speed keeps growing, and by t = 15 s (when B's speed equals A's 30 m/s) the gap has already reached its minimum and starts widening again. Always check the discriminant before solving for t.

JEE AdvancedVelocity-proportional decelerationDifferential-equation kinematics

A boat moving at initial speed $v_{0}$ experiences water resistance such that its acceleration is $a = - k v$ , where k is a positive constant. Show that the boat's velocity decays exponentially with time, and that despite never reaching exactly zero in finite time, it travels only a finite total distance.

Stuck? Show the approach

This acceleration is not constant, so none of the three standard kinematic equations apply. Go back to $a = d v / d t$ , separate variables, and integrate directly — this is exactly the calculus route the chapter derives the standard equations from, applied to a case where a is not fixed.

Show the full solution

$\frac{d v}{d t} = - k v ⟹ \int_{v_{0}}^{v} \frac{d v}{v} = - k \int_{0}^{t} d t ⟹ ln (\frac{v}{v _{0}}) = - k t ⟹ v = v_{0} e^{- k t}$

Total distance: $x = \int_{0}^{\infty} v_{0} e^{- k t} d t = \frac{v _{0}}{k} [- e^{- k t}]_{0}^{\infty} = \frac{v _{0}}{k} (0 - (- 1)) = \frac{v _{0}}{k}$

Answer: v = v0 e^(-kt); total distance travelled = v0/k, a finite value

The trap

It seems paradoxical that a boat which (mathematically) never fully stops still covers only a finite distance — but this is exactly what an exponential decay does: it gets closer to zero speed fast enough that the area under its v-t curve (the total distance) still converges to a finite number, v0/k.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Kinematics)High

NEET PhysicsHigh

NSEP and Indian National Physics OlympiadMedium

State engineering entrance papers (MHT-CET, WBJEE, KCET)High

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Questions students ask

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It genuinely is not there. The 2026-27 NCERT chapter never uses the word 'frame' at all — the closest it comes is telling you to measure position 'with reference to a conveniently chosen origin' and to pick a positive direction. But CBSE does list it as examinable for this chapter, so you need it from somewhere. A frame of reference is simply the coordinate system an observer uses: an origin, a positive direction, and a clock. Its importance is that motion is meaningless without one — a passenger sitting on a moving train is at rest in the train's frame and moving at 80 km/h in the platform's frame, and neither answer is wrong. This is also the idea underneath relative velocity, where v_AB = v_A minus v_B is the rule for converting a velocity from one frame to another.

Not in the 2026-27 reprint. The contents page still lists a section 2.5 called Relative velocity, and the introduction still promises to introduce it, but read the chapter body end to end and nothing appears between the kinematic equations and the Summary — no definition, no formula, no worked example. The label was left behind when the content was cut. This matters because Exercise 2.14, the police van firing at a fleeing car, is a relative-velocity problem in everything but name. The chapter notes cover exactly what that exercise needs: v_AB = v_A minus v_B as signed quantities, speeds subtracting for same-direction motion and adding for opposite directions.

Because 'the limit as delta-t approaches zero' is a phrase that is easy to recite and hard to actually believe. The chapter takes windows of 2.0 s, 1.0 s, 0.5 s, 0.1 s and 0.01 s, all centred on t = 4 s, and shows the average velocity marching steadily toward 3.84 m/s. By the last row the number has stopped moving in any digit that matters. That is the whole point: you can watch the limiting process converge instead of taking it on faith, and the value it converges to is precisely dx/dt at that instant.

This was genuinely an open question in Galileo's day, and the chapter says so. It was first thought that change of velocity with distance might be the natural definition. Galileo's studies of freely falling bodies and of objects on inclined planes settled it: the rate of change of velocity with time is constant for all objects in free fall, while the rate of change with distance is not constant at all — it decreases as the fall continues. Defining acceleration against time therefore produces a quantity that stays fixed in the most important case, which is what makes it useful.

Look at the units on the two axes rather than at the picture. The vertical axis is velocity in m/s, the horizontal axis is time in s, so multiplying a height by a width gives (m/s) times (s), and the seconds cancel to leave metres. The area is not a distance by analogy — it is a distance dimensionally. For the simplest case, an object moving at constant velocity u for time T, the area under the flat v-t line is the rectangle u times T, which is exactly the displacement. The general result for any v-t curve needs calculus, but the reason is the same.

The second one. The first method splits the flight into an upward part and a downward part and computes each time separately, which works but requires you to reason correctly about the path. The second method just writes down the initial position, final position, initial velocity and acceleration, and solves a single quadratic in t. The chapter states plainly that the second is better, because the kinematic equations only need the start and end states under constant acceleration — they are indifferent to what happened in between. Fewer steps means fewer places to make a sign error.

Yes, and this is one of the most commonly missed points in the chapter. Velocity passes through zero at the highest point, but acceleration does not — gravity acts continuously throughout the flight, before, during and after that instant. The acceleration-time graph for the whole flight is a flat horizontal line at minus g with no break anywhere in it. If acceleration really did vanish at the top, the ball would have no reason to start falling and would simply stay suspended there. Points to Ponder #4 states this directly.

Because stopping distance goes as the square of the initial speed, not as the speed itself. Setting final velocity to zero in v^2 = v0^2 + 2ax gives d_s = v0^2 / 2a, so doubling v0 multiplies the distance by four for the same braking deceleration. The chapter backs this with measured data from a real car: braking distances of 10, 20, 34 and 50 m at speeds of 11, 15, 20 and 25 m/s. That square law, combined with reaction time from Example 2.7, is the whole physics behind why speed limits near schools are set as low as they are.

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