Class 9 — Motion Deep Dive

Chapter Overview & Weightage

Motion is the very first physics chapter in Class 9 and sets the foundation for all of mechanics. In CBSE Class 9 Science papers, this chapter consistently fetches $4$ – $6$ marks — usually one short-answer question on definitions and one numerical on equations of motion.

CBSE Class 9 Science — Motion Weightage

Year	Marks from Motion	Type of Questions
2024	$5$	$1$ definition, $1$ numerical, $1$ graph
2023	$6$	$1$ derivation, $2$ numericals
2022	$4$	$1$ short, $1$ numerical
2021	$5$	$1$ MCQ, $1$ numerical, $1$ graph
2020	$5$	Mixed

The chapter also seeds Class 11 Kinematics — students who get this right find Class 11 much smoother.

The skills you build here — distance vs displacement, sketching motion graphs, applying $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ — are tested every year. Treat this as a scoring topic, not a hard one.

Key Concepts You Must Know

In rough order of how often they appear in CBSE Class 9 papers:

Distance vs Displacement (highest priority)

Distance: total path length; scalar; always $\geq 0$ .
Displacement: shortest line from start to end; vector; can be zero.
Classic question: a body covers a semicircular arc of radius $R$ . Distance $= \pi R$ , displacement $= 2R$ (the diameter).

Speed vs Velocity

Speed = distance/time; scalar.
Velocity = displacement/time; vector. Sign matters!

Uniform vs Non-uniform Motion

Uniform: equal distances in equal intervals (like a metro train cruising at constant speed).
Non-uniform: changing speed or direction.

Acceleration

Rate of change of velocity. Unit: m/s². If velocity decreases, acceleration is negative (deceleration).

Equations of Motion (constant acceleration)

$v = u + at$
$s = ut + \tfrac{1}{2}at^2$
$v^2 = u^2 + 2as$

Motion Graphs (very common)

Distance-time slope = speed.
Velocity-time slope = acceleration; area = displacement.

Circular Motion

Even at constant speed, direction changes → there is acceleration (centripetal).

Important Formulas

v = u + at

s = ut + \tfrac{1}{2}at^2

v^2 = u^2 + 2as

When to use: any problem with constant acceleration. Identify which $4$ of $\{u, v, a, s, t\}$ you have. The equation that uses those $4$ is the right one.

\text{Average speed} = \tfrac{\text{total distance}}{\text{total time}}

\text{Average velocity} = \tfrac{\text{total displacement}}{\text{total time}}

For uniform acceleration only: $\bar{v} = \tfrac{u+v}{2}$ .

When to use: when the question asks for average over a journey, especially with multiple legs at different speeds.

s_n = u + \tfrac{a(2n-1)}{2}

The distance covered in the $n^{\text{th}}$ second alone (not in $n$ seconds).

When to use: specific CBSE-favourite question type, e.g., “distance covered in the $5^{\text{th}}$ second”. Don’t use $s = ut + \tfrac{1}{2}at^2$ here — that gives total distance.

Solved Previous Year Questions

PYQ 1 — Definitions and Units (CBSE 2023, $3$ marks)

Define acceleration. Give its SI unit. State the type of acceleration when (a) speed is constant and direction changes, (b) speed decreases.

Answer: Acceleration is the rate of change of velocity with time. SI unit: $\text{m/s}^2$ .

(a) Centripetal acceleration (toward the centre). (b) Negative acceleration (also called deceleration or retardation).

PYQ 2 — Equations of Motion (CBSE 2024, $3$ marks)

A train starts from rest and acquires a velocity of $72$ km/h in $5$ minutes. Calculate (a) the acceleration, (b) the distance covered.

Solution:

Convert: $72$ km/h $= 72 \times \tfrac{1000}{3600} = 20$ m/s. $5$ min $= 300$ s.

(a) $v = u + at \implies 20 = 0 + a \times 300 \implies a = \tfrac{1}{15}$ m/s² $\approx 0.067$ m/s².

(b) $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times \tfrac{1}{15} \times 300^2 = \tfrac{1}{2} \times \tfrac{1}{15} \times 90000 = 3000$ m $= 3$ km.

PYQ 3 — Distance vs Displacement (CBSE 2022, $5$ marks)

A cyclist travels $4$ km east, then $3$ km north. Find the distance covered and the magnitude of displacement.

Solution:

Distance: $4 + 3 = 7$ km.

Displacement: vector from start to end forms a right triangle with legs $4$ km and $3$ km. Magnitude $= \sqrt{16 + 9} = 5$ km. Direction: $\tan^{-1}(3/4)$ north of east.

Difficulty Distribution

Based on CBSE Class 9 papers from $2020$ – $2024$ :

Difficulty	% of Motion Questions	Question Types
Easy	$40\%$	Definitions, MCQs on units, simple graph reading
Medium	$50\%$	Direct application of equations of motion, distance/displacement word problems
Hard	$10\%$	Multi-stage motion (acceleration then deceleration), $n^{\text{th}}$ second formula

The “hard” $10\%$ usually appears as the long-answer numerical worth $4$ – $5$ marks. Practise $5$ – $6$ multi-stage motion problems before the exam.

Expert Strategy

Week 1 — Build vocabulary: Make flashcards for distance, displacement, speed, velocity, acceleration, uniform/non-uniform motion. Drill until definitions are reflexive.

Week 2 — Master the three equations: Practise identifying which equation to use. The trick: count knowns. If $t$ is missing, use $v^2 = u^2 + 2as$ . If $v$ (final) is missing, use $s = ut + \tfrac{1}{2}at^2$ .

Week 3 — Graph fluency: Draw d-t and v-t graphs for $5$ different scenarios: uniform motion, uniform acceleration from rest, uniform deceleration, free fall, and a stop-go journey. Read off slopes and areas.

Topper’s habit: every time you write a numerical, write units in every line — not just the final answer. Boards reward this; it earns the “process” mark even if arithmetic slips.

Last-week revision: Solve all NCERT exemplar problems for Motion. Then do CBSE PYQs from $2018$ – $2024$ — that’s $\sim 12$ questions, doable in $4$ hours total.

Common Traps

Trap 1: Confusing distance and displacement when the body returns. A boy walks $5$ m east and $5$ m back west — distance is $10$ m, displacement is $0$ . Many students write $10$ for both.

Trap 2: Using $g = 9.8$ when the question specifies $g = 10$ . Read the problem statement carefully. CBSE Class 9 usually uses $g = 10$ for arithmetic ease, but exam directions matter.

Trap 3: Mixing up units. Speed in km/h, time in seconds, asking for distance in metres. Convert all quantities to a single unit system before plugging in. The most common trap: $72$ km/h treated as $72$ m/s.

Trap 4: Sign of acceleration. Deceleration (a body slowing down) means $a$ is negative when motion is in the positive direction. Forgetting the sign in $v^2 = u^2 + 2as$ gives a meaningless negative under the square root.

Trap 5: $s_n$ vs $s$ . Distance in the $n^{\text{th}}$ second is $s_n = u + \tfrac{a(2n-1)}{2}$ , which is different from total distance in $n$ seconds, $s = ut + \tfrac{1}{2}at^2$ . CBSE board examiners often pose this trap on purpose.

In CBSE 2024, a $3$ -marker asked for distance covered in the $4^{\text{th}}$ second of a body starting from rest with $a = 2$ m/s². Answer using $s_n$ formula: $s_4 = 0 + (2 \times 7)/2 = 7$ m. Many students wrote $16$ m using total-distance formula — and lost all $3$ marks.

Quick Revision Card

Distance is scalar; displacement is vector with magnitude and direction.
Equations of motion apply only when acceleration is constant.
Slope of v-t graph = acceleration; area = displacement.
Centripetal acceleration is directed toward the centre, even when speed is constant.
$s_n \neq s$ . Don’t confuse the two.

Master these five lines and you’ve got $90\%$ of the chapter’s marks.