Chapter 7 of 12

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CBSE Class 9 Science★ 8–10 marks · CBSE Class 9 board (Unit III — Motion, Force and Work)

Motion

How fast? In what direction? How far? Class 9 Motion covers reference frames, distance vs displacement, speed vs velocity, acceleration, three equations of motion derived graphically, uniform circular motion, and 25+ exam-style numericals with full step-by-step solutions and worked graph problems.

⏱ 45 min readMedium difficulty✓ Free · NCERT-aligned

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

1Distinguish scalar and vector quantities; classify distance, displacement, speed, velocity, acceleration
2Convert speeds between km/h and m/s using the 18/5 factor
3Plot and interpret distance-time and velocity-time graphs
4Derive the three equations of motion graphically (v-t graph area method)
5Apply the equations v = u + at, s = ut + ½at², v² = u² + 2as to numerical problems
6Identify uniform circular motion and explain why acceleration is non-zero despite constant speed
7Compute average speed and average velocity from compound motion problems

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Why this chapter matters

Motion is the gateway to mechanics — every physics problem you'll solve from now to engineering college builds on these three equations. The distance vs displacement and speed vs velocity distinctions reappear in every vector problem you'll ever do.

Before you start — revise these

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

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Algebra (Class 8)

Solving linear and quadratic equations.

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Cartesian coordinates (Class 9 Math)

Reading graphs and slopes.

Motion — Class 9 (CBSE)

Open a window of a moving train and look out: trees seem to fly backward, the moon stands still, and a distant car overtakes you. Whose motion is which? This chapter is about how physicists make sense of "what's moving and how fast" — and how they distill it into three simple equations that describe everything from a falling apple to a rocket leaving Earth.

1. The story — motion as the simplest physics

In 1638, Galileo Galilei published Two New Sciences. It contained the first quantitative treatment of motion: how a ball rolls down a ramp, accelerates uniformly, and follows precisely-predictable mathematics. This was the birth of physics.

Before Galileo, Aristotle's view dominated for 2000 years: heavy objects fall faster than light ones; objects naturally come to rest; motion requires a continuous push. Galileo (and later Newton) overturned all of this.

This chapter is foundational. The three equations of motion you'll derive here describe:

A ball you throw upward.
A bullet fired from a gun.
A car braking to a stop.
A satellite circling the Earth.
A planet orbiting the sun (with one tweak).

Master them and you have the language of mechanics.

2. Reference frame — motion is relative

A passenger sits in a moving train. To another passenger, she's at rest. To someone on the platform, she's moving at 60 km/h. Both are right.

A reference frame is the object or location you use to measure positions and velocities. There's no "absolute" rest in the universe — motion is always relative to a chosen reference frame.

For Class 9 problems, the reference frame is usually the ground (the Earth), but specify it explicitly if you're working with two moving objects.

3. Distance vs displacement

These two are the most-confused concepts in this chapter. Don't.

Distance = the total path length traveled. Scalar (only magnitude). Always positive.

Displacement = the shortest straight-line distance from initial to final position, with direction. Vector (magnitude + direction). Can be positive, negative, or zero.

Example: You walk 5 m east, then 5 m back west, returning to your start.

Distance: 5 + 5 = 10 m.
Displacement: 0 (same start and end point).

Example: You walk 3 m east, then 4 m north.

Distance: 3 + 4 = 7 m.
Displacement: $3^{2} + 4^{2} = 5$ m, northeast (specifically at $tan^{- 1} (4/3) \approx 53°$ north of east).

Key contrasts

Feature	Distance	Displacement
Type	Scalar	Vector
Sign	Always positive	Can be 0, +, or −
Path	Whole path	Straight line, start to end
Magnitude	≥ displacement	≤ distance

The "magnitude of displacement ≤ distance" inequality is a 1-mark factoid. They're equal only when the motion is in a straight line without reversing.

4. Speed vs velocity

Built on the previous distinction.

Speed = distance per unit time. Scalar. $Speed = \frac{distance}{time}$

Velocity = displacement per unit time. Vector. $Velocity = \frac{displacement}{time}$

SI unit of both: m/s (also commonly km/h).

Conversion: $1 km/h = \frac{5}{18} m/s$ , or equivalently $1 m/s = \frac{18}{5} km/h = 3.6 km/h$ .

Uniform vs non-uniform

Uniform motion: the body covers equal distances in equal intervals of time. Speed is constant.
Non-uniform motion: distances covered vary with time. Speed is changing.

Average vs instantaneous

Average speed = total distance / total time.
Average velocity = total displacement / total time.
Instantaneous speed/velocity = the value at one specific instant.

5. Acceleration — the rate of change of velocity

If velocity changes (in magnitude OR direction), the object is accelerating.

$a = \frac{v - u}{t}$

Where:

$u$ = initial velocity (m/s)
$v$ = final velocity (m/s)
$t$ = time interval (s)
$a$ = acceleration (m/s²)

Three cases

Positive acceleration ( $a > 0$ , velocity increasing): a car speeding up.
Negative acceleration / deceleration / retardation ( $a < 0$ , velocity decreasing): a car braking.
Zero acceleration ( $a = 0$ , constant velocity): a car cruising at 60 km/h on a flat road.

SI unit of acceleration: m/s².

Uniform vs non-uniform acceleration

Uniform: constant value (e.g., free fall has $a = g = 9.8 m/s^{2}$ , constant).
Non-uniform: changing (e.g., a car accelerating then decelerating in traffic).

6. Motion graphs

Distance-time (or position-time) graph

Slope = speed.
Horizontal line = at rest (no distance change).
Straight line going up = uniform speed.
Curve (slope changing) = changing speed (non-uniform motion).

Velocity-time graph

Slope = acceleration.
Horizontal line = constant velocity (zero acceleration).
Straight line going up = uniform acceleration.
AREA under the curve = displacement (very important for derivations and numericals).

Both graphs are powerful tools that show motion at a glance — and on the CBSE exam, frequently appear with a 3–4 mark question asking you to compute speed or displacement from them.

7. The three equations of motion (uniform acceleration)

These three equations describe motion when acceleration is constant (which covers free fall, braking, and many CBSE problems).

$v = u + a t$ $s = u t + \frac{1}{2} a t^{2}$ $v^{2} = u^{2} + 2 a s$

Where $s$ is displacement.

Derivation — equation 1

Acceleration: $a = (v - u) / t$ , so $a t = v - u$ , so $v = u + a t$ .

Derivation — equation 2 (graphically)

Plot a $v$ – $t$ graph: starts at $u$ , ends at $v$ at time $t$ , straight line in between.

Displacement = area under this graph = area of a trapezium = $\frac{1}{2} (u + v) t$ . Substitute $v = u + a t$ : $s = \frac{1}{2} (u + u + a t) t = u t + \frac{1}{2} a t^{2}$

Derivation — equation 3

Square equation 1: $v^{2} = (u + a t)^{2} = u^{2} + 2 (u) (a t) + (a t)^{2}$ . Then $v^{2} - u^{2} = 2 u (a t) + a^{2} t^{2}$ . The right-hand side is $2 a (u t + \frac{1}{2} a t^{2}) = 2 a s$ . So $v^{2} = u^{2} + 2 a s$ .

How to pick which equation

Read the problem and check what's GIVEN and what's WANTED:

Don't have time? Use the one WITHOUT $t$ : $v^{2} = u^{2} + 2 a s$ .
Don't have $v$ ? Use the one WITHOUT $v$ : $s = u t + \frac{1}{2} a t^{2}$ .
Don't have $s$ ? Use the one WITHOUT $s$ : $v = u + a t$ .

This three-equation decision tree is everything for solving Class 9 motion numericals.

8. Uniform circular motion — a special case

An object moving along a circle at constant speed is in uniform circular motion. Examples: a satellite orbiting Earth, a stone tied to a string twirled at constant speed, the moon around Earth.

Even though speed is constant, velocity is changing — because direction is changing every instant. Hence the object IS accelerating (changing velocity = acceleration). The acceleration always points toward the centre of the circle and is called centripetal acceleration.

For an object moving at speed $v$ on a circle of radius $r$ :

Speed = $\frac{2 π r}{T}$ where $T$ is the time period.
Centripetal acceleration $= v^{2} / r$ .

You'll derive the centripetal-force formula in Class 11 — for Class 9, just know:

Speed constant ≠ acceleration zero in circular motion.
The force keeping the object on the circle is centripetal force, directed toward the centre.

9. Worked example — a complete walkthrough

Question: A car starts from rest and accelerates uniformly at $2 m/s^{2}$ for 5 s. Find (a) final velocity, (b) displacement in this time, (c) using a different equation, verify the displacement.

Step 1 — Note givens. $u = 0$ , $a = 2 m/s^{2}$ , $t = 5 s$ .

Step 2 — Find $v$ . Use $v = u + a t = 0 + 2 (5) = 10 m/s$ .

Step 3 — Find $s$ . Use $s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} (2) (25) = 25 m$ .

Step 4 — Verify using $v^{2} = u^{2} + 2 a s$ . $1 0^{2} = 0 + 2 (2) (s) \Rightarrow s = 100/4 = 25 m$ . ✓

10. Closing thought

Three equations. They describe everything that moves in a straight line under constant acceleration. They power the calculations behind:

Stopping distances on roads (with friction giving deceleration).
Free-fall problems (with $a = g = 9.8 m/s^{2}$ downward).
Projectile motion (Class 11 — horizontal + vertical decomposed).
Rocket launches (with $a$ changing as fuel burns).
Sports physics (a sprinter's start, a footballer's kick, a cricketer's throw).

A century before computers, generations of engineers built bridges, ships and aircraft using exactly these equations. Today they're hardcoded into every physics simulation, every video-game engine, every flight simulator. Master them and you've mastered the alphabet of physics.

Key formulas & results

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

Speed

speed = distance / time

Scalar. SI unit m/s. 1 km/h = 5/18 m/s.

Velocity

velocity = displacement / time

Vector. Same units as speed but with direction.

Average speed

= total distance / total time

NOT the average of two speeds — must use total distance and total time.

Acceleration

a = (v − u) / t

v = final velocity, u = initial, t = time. SI m/s².

1st equation of motion

v = u + at

Use when: need v or t and have u, a.

2nd equation of motion

s = ut + ½at²

Use when: need s (displacement) and have u, a, t. NO v needed.

3rd equation of motion

v² = u² + 2as

Use when: NO t given. Have u, v, a, or s.

Area under v-t graph

Area = displacement (s)

Single most-useful graphical fact.

Slope of v-t graph

= acceleration

Constant slope = uniform acceleration.

Slope of s-t graph

= speed (or velocity, if signed)

Steeper slope = faster motion.

Speed in circular motion

v = 2πr / T

r = radius, T = time period. Circle's circumference / time.

Centripetal acceleration

a = v² / r

Directed toward centre. Class 11 derivation; Class 9 just states it.

⚠️

Common mistakes & fixes

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

WATCH OUT

✗ Confusing distance and displacement

✓ Distance = scalar, total path. Displacement = vector, shortest line from start to end. Magnitude of displacement ≤ distance always.

WATCH OUT

✗ Computing average speed as (v₁ + v₂) / 2

✓ WRONG unless times are equal. Always use: avg speed = total distance / total time. For a journey split into two equal distances, average = 2v₁v₂ / (v₁ + v₂) (harmonic mean).

WATCH OUT

✗ Forgetting to convert km/h to m/s

✓ Always check units. If a is in m/s², all velocities must be in m/s. 1 km/h = 5/18 m/s.

WATCH OUT

✗ Treating negative velocity as 'wrong' direction without context

✓ Negative velocity means motion in the −ve direction of the chosen axis. Always SPECIFY the positive direction explicitly.

WATCH OUT

✗ Saying uniform circular motion has zero acceleration

✓ Speed constant ≠ velocity constant. In circular motion, direction is changing → velocity changing → acceleration NONZERO (centripetal).

WATCH OUT

✗ Choosing the wrong equation for the numerical

✓ First list givens (u, v, a, t, s). Then pick the equation that has all your knowns + one unknown. Equations differ in which variable they EXCLUDE.

WATCH OUT

✗ Using g = +9.8 for objects moving upward

✓ g acts DOWNWARD always. For upward motion, take it as −9.8 m/s² if up is positive; +9.8 if down is positive. Set sign by your axis choice.

NCERT exercises (with solutions)

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

Section 7.1 (in-text)

Reference frames, rest vs motion

Questions

Section 7.2 (in-text)

Distance vs displacement, scalar vs vector

Questions

Section 7.3 (in-text)

Speed, velocity, uniform vs non-uniform motion

Questions

Section 7.4 (in-text)

Acceleration; deceleration; numericals

Questions

Section 7.5 (in-text)

Graphical representation of motion; s-t and v-t graphs

Questions

Section 7.6 (in-text)

Three equations of motion: derivations + problems

Questions

Section 7.7 (in-text)

Uniform circular motion

Questions

End-of-chapter

Mixed: conversions, average speed, equations of motion, graph reading

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Distance/disp

A person walks 4 m east, then 3 m north. Find the distance and magnitude of displacement.

▶ Show solution

Step 1 — Distance = 4 + 3 = 7 m.

Step 2 — Displacement: a right-triangle from start to end. Magnitude = √(4² + 3²) = √(16 + 9) = √25 = 5 m.

✦ Answer: Distance = 7 m, displacement (magnitude) = 5 m (toward NE, at 37° N of E).

Q2EASY· Unit

Convert 72 km/h to m/s.

▶ Show solution

Step 1 — Use 1 km/h = 5/18 m/s.

Step 2 — 72 × 5/18 = 360/18 = 20 m/s.

✦ Answer: 20 m/s.

Q3EASY· Eqn 1

A car starts from rest and acquires a velocity of 20 m/s in 4 s. Find its acceleration.

▶ Show solution

Step 1 — Identify givens.
  u = 0 (starts from rest), v = 20 m/s, t = 4 s.

Step 2 — Use a = (v − u)/t.
  a = (20 − 0)/4 = 5 m/s².

✦ Answer: 5 m/s².

Q4EASY· Eqn 2

A car moving at 10 m/s accelerates at 2 m/s² for 5 s. Find the distance covered.

▶ Show solution

Step 1 — Givens: u = 10 m/s, a = 2 m/s², t = 5 s. Unknown s, no v needed → use eqn 2.

Step 2 — s = ut + ½at² = 10(5) + ½(2)(5)² = 50 + 25 = 75 m.

✦ Answer: 75 m.

Q5EASY· Concept

State the SI units of (a) displacement, (b) velocity, (c) acceleration.

▶ Show solution

Step 1 — Recall the basic SI units.
  (a) Displacement: metre (m). Same unit as distance.
  (b) Velocity: metres per second (m/s). 
  (c) Acceleration: metres per second per second (m/s²).

✦ Answer: (a) m, (b) m/s, (c) m/s².

Q6MEDIUM· Eqn 3

A bullet leaves a 50-cm-long rifle barrel with a velocity of 400 m/s. Find its acceleration assuming uniform acceleration in the barrel.

▶ Show solution

Step 1 — Identify givens.
  u = 0 (starts from rest in the barrel), v = 400 m/s, s = 50 cm = 0.5 m.
  No t given → use 3rd equation.

Step 2 — v² = u² + 2as → 400² = 0 + 2 × a × 0.5.

Step 3 — Solve.
  160000 = a → a = 160000 m/s² = 1.6 × 10⁵ m/s².

✦ Answer: a ≈ 1.6 × 10⁵ m/s².

For comparison, gravity is ~10 m/s². The bullet's acceleration is ~ 16,000 times stronger — that's why rifles produce muzzle velocities of hundreds of m/s in fractions of a second.

Q7MEDIUM· Avg speed

A train travels half its journey at 60 km/h and the other half at 40 km/h. Find the average speed.

▶ Show solution

Step 1 — Let total distance be 2d.
  First half (d) at 60 km/h: time t₁ = d/60.
  Second half (d) at 40 km/h: time t₂ = d/40.

Step 2 — Total time.
  t = t₁ + t₂ = d/60 + d/40 = (4d + 6d)/240 = 10d/240 = d/24.

Step 3 — Average speed.
  v_avg = total distance / total time = 2d / (d/24) = 2d × 24/d = 48 km/h.

✦ Answer: 48 km/h.

Note: NOT (60+40)/2 = 50. Average speed uses harmonic mean when distances are equal:
   v_avg = 2v₁v₂ / (v₁ + v₂) = 2 × 60 × 40 / (60 + 40) = 4800/100 = 48 km/h. ✓

Q8MEDIUM· Free fall

A stone is dropped from a height of 45 m. Find (a) the time to reach the ground, (b) the velocity just before hitting the ground. (g = 10 m/s²)

▶ Show solution

Step 1 — Givens.
  u = 0 (dropped, not thrown), a = g = 10 m/s² (downward), s = 45 m (downward; take downward positive).

Step 2 (a) — Find time.
  Use s = ut + ½at² ⇒ 45 = 0 + ½ × 10 × t² ⇒ t² = 9 ⇒ t = 3 s.

Step 3 (b) — Final velocity.
  Use v = u + at = 0 + 10 × 3 = 30 m/s.
  OR v² = u² + 2as = 0 + 2 × 10 × 45 = 900 ⇒ v = 30 m/s. ✓

✦ Answer: (a) 3 s; (b) 30 m/s.

Q9MEDIUM· Stopping

A car moving at 36 km/h is brought to rest by applying brakes over a distance of 10 m. Find the deceleration produced.

▶ Show solution

Step 1 — Convert speed.
  36 km/h = 36 × 5/18 = 10 m/s.

Step 2 — Givens.
  u = 10 m/s, v = 0, s = 10 m. Use v² = u² + 2as.

Step 3 — Substitute.
  0² = 10² + 2 × a × 10
  0 = 100 + 20a
  a = −5 m/s².

✦ Answer: Deceleration = 5 m/s² (negative sign indicates motion is decelerating).

Real-world: braking decelerations on dry road are typically 3–7 m/s². Wet road or worn tyres reduce this.

Q10MEDIUM· Throw up

A ball is thrown vertically upward with initial velocity 20 m/s. (a) How high does it rise? (b) How long before it hits the ground? (g = 10 m/s²)

▶ Show solution

Step 1 — Pick up as positive. Givens: u = +20, a = −10 m/s² (gravity opposes upward motion).

Step 2 (a) — At max height, v = 0. Use v² = u² + 2as.
  0 = 400 + 2(−10)s
  s = 400/20 = 20 m.

Step 3 (b) — Total time = time to reach max height + time to fall back (symmetric, no air resistance).
  Time up: v = u + at ⇒ 0 = 20 − 10t ⇒ t_up = 2 s.
  By symmetry, time down = 2 s.
  Total = 4 s.

✦ Answer: (a) Maximum height = 20 m. (b) Total time in the air = 4 s.

Q11HARD· Graph

On a velocity-time graph, a body's velocity increases linearly from 0 to 20 m/s in 10 s, stays constant for 5 s, then decreases linearly to 0 in 5 s. Find: (a) acceleration in each phase, (b) total distance covered.

▶ Show solution

Step 1 — Phase 1 (0 to 10 s, velocity 0 → 20 m/s):
  a₁ = (20 − 0)/10 = 2 m/s².

Step 2 — Phase 2 (10 to 15 s, velocity constant at 20 m/s):
  a₂ = 0.

Step 3 — Phase 3 (15 to 20 s, velocity 20 → 0):
  a₃ = (0 − 20)/5 = −4 m/s² (deceleration of 4 m/s²).

Step 4 — Total distance = total area under v-t graph.
  Phase 1: triangle of base 10, height 20 → area = ½ × 10 × 20 = 100 m.
  Phase 2: rectangle of width 5, height 20 → 5 × 20 = 100 m.
  Phase 3: triangle of base 5, height 20 → ½ × 5 × 20 = 50 m.
  Total = 100 + 100 + 50 = 250 m.

✦ Answer: (a) Phase 1: 2 m/s²; Phase 2: 0; Phase 3: −4 m/s² (deceleration 4 m/s²). (b) Total distance = 250 m.

Q12HARD· Two body

Two cars start from the same point at the same time. Car A moves with constant velocity 20 m/s. Car B starts from rest with acceleration 2 m/s². At what time will they meet again?

▶ Show solution

Step 1 — Set up equations for position.
  Car A: x_A = 20t (uniform motion).
  Car B: x_B = 0 × t + ½ × 2 × t² = t² (uniform acceleration from rest).

Step 2 — They meet again when x_A = x_B.
  20t = t² ⇒ t² − 20t = 0 ⇒ t(t − 20) = 0.
  Solutions: t = 0 (initial start) or t = 20 s.

Step 3 — Meaningful answer.
  They meet again at t = 20 s.
  Distance from start: x = 20 × 20 = 400 m (or t² = 400 m ✓).

✦ Answer: They meet again at t = 20 s, at a distance of 400 m from the start. Car B was slower initially but kept accelerating, so it eventually caught up.

Q13HARD· Falling

A stone is dropped from a tower. It covers 25 m in the LAST second of its fall. Find the height of the tower. (g = 10 m/s²)

▶ Show solution

Step 1 — Let total time of fall be n seconds.
  Distance fallen in n seconds: S_n = ½ × g × n² = 5n².
  Distance fallen in (n − 1) seconds: S_{n−1} = 5(n − 1)².

Step 2 — Distance covered in the last second = S_n − S_{n−1}.
  = 5n² − 5(n − 1)²
  = 5n² − 5(n² − 2n + 1)
  = 5(2n − 1)
  = 10n − 5.

Step 3 — Set equal to 25 m.
  10n − 5 = 25 ⇒ 10n = 30 ⇒ n = 3 s.

Step 4 — Total height = S_n = 5n² = 5 × 9 = 45 m.

✦ Answer: Tower height = 45 m. Fall time = 3 s.

Sanity check: Distance in 2 s = 5 × 4 = 20 m. Distance in 3 s = 45 m. Difference = 25 m ✓.

Q14HARD· Circular

The Earth orbits the Sun in a roughly circular path of radius 1.5 × 10⁸ km. It takes 365.25 days. Find the orbital speed of the Earth in km/s.

▶ Show solution

Step 1 — Use v = 2πr/T.
  r = 1.5 × 10⁸ km.
  T = 365.25 days = 365.25 × 24 × 60 × 60 s = 3.156 × 10⁷ s.

Step 2 — Plug in.
  v = 2π × 1.5 × 10⁸ / 3.156 × 10⁷
   = 9.425 × 10⁸ / 3.156 × 10⁷
   ≈ 29.86 km/s.

✦ Answer: Earth's orbital speed ≈ 30 km/s.

That's about 108,000 km/h. We're hurtling through space at this speed without feeling anything, because we're all moving together.

Q15HARD· HOTS

A bus accelerates uniformly from rest and covers 36 m in the 4th second. Find (a) the acceleration, (b) the velocity at the end of the 4th second.

▶ Show solution

Step 1 — Use the 'distance covered in nth second' formula derived from equations of motion.
  S_n = u + a(n − 1/2) = u + ½ × a × (2n − 1).
  For u = 0: S_n = ½ × a × (2n − 1).

Step 2 — Substitute.
  For n = 4: S_4 = ½ × a × (2 × 4 − 1) = ½ × a × 7 = 3.5a.

Step 3 — Set S_4 = 36 m.
  3.5a = 36 ⇒ a = 36/3.5 ≈ 10.29 m/s².
  (Or exactly: a = 72/7 m/s².)

Step 4 — Velocity at end of 4 s.
  v = u + at = 0 + 10.29 × 4 ≈ 41.14 m/s.
  (Or exactly: 288/7 m/s.)

✦ Answer: (a) a ≈ 10.29 m/s²; (b) v ≈ 41.14 m/s at the end of the 4th second.

The 'nth second' formula S_n = u + ½a(2n−1) is a clever shortcut — derived from S_n = S(n) − S(n−1) where S(t) = ut + ½at². Worth memorising.

5-minute revision

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

•Motion is relative to a reference frame. There's no absolute rest.
•Distance = scalar (always positive, full path). Displacement = vector (start-to-end straight line, can be 0).
•Magnitude of displacement ≤ distance, equality only for straight-line motion without reversal.
•Speed = scalar = distance/time. Velocity = vector = displacement/time. SI units: m/s.
•Conversion: 1 km/h = 5/18 m/s.
•Average speed = total distance / total time. For equal distances at two speeds, harmonic mean: 2v₁v₂/(v₁+v₂).
•Acceleration = (v − u) / t. SI unit: m/s². Negative a = deceleration.
•Three equations (uniform acceleration): v = u + at; s = ut + ½at²; v² = u² + 2as.
•Choose the equation that lacks the variable you don't know.
•s-t graph slope = speed. v-t graph slope = acceleration. v-t graph AREA = displacement.
•Uniform circular motion: speed constant, velocity NOT (direction changes). Acceleration is centripetal, directed inward.
•Free-fall: u = 0 if dropped, a = g = 9.8 m/s² (often rounded to 10).

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Displacement encodes both how far AND in which direction you went. Distance just measures the total path traversed. Two opposite vectors of equal magnitude cancel; two scalars never cancel. That's the whole vector-vs-scalar distinction.

No — speed is scalar and always ≥ 0. Only VELOCITY can be negative (motion in the −ve direction of your chosen axis). Newcomers often write 'negative speed' but should say 'negative velocity' or 'deceleration'.

v-t graph: y-axis is velocity (m/s), x-axis is time (s). Their product (area) has units m/s × s = m — which is distance/displacement. Geometrically, breaking the graph into tiny rectangles of width dt gives velocity × dt = small displacements, summing to total displacement.

Deceleration just means the velocity's magnitude is DECREASING. If you've set forward as positive, the velocity is positive but acceleration is negative. The signs follow your axis convention, not the direction of motion in space.

Direction is changing every instant. Velocity (a vector) is changing. Velocity changing = acceleration. Acceleration requires force (Newton's 2nd law). That force is gravity, pulling the satellite continuously toward Earth's centre. Without it, the satellite would fly off in a straight line.

Yes — average velocity's magnitude can be less than average speed if the path isn't straight. Example: walk 5 m east then 5 m west in 10 s. Average speed = 10/10 = 1 m/s. Average velocity = 0 (zero displacement).

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Matter in Our Surroundings

From the air you breathe to the ice in your glass — everything around you is matter. Class 9 Matter in Our Surroundings with the particle theory, the three classical states, the curious fourth and fifth states (plasma and Bose-Einstein condensate), latent heat, evaporation, and 25+ exam-style problems with full step-by-step solutions.

Is Matter Around Us Pure?

Almost nothing in everyday life is chemically "pure" — your air is a mixture, your milk is a colloid, your salt water is a solution. Class 9 Is Matter Around Us Pure? distinguishes pure substances, mixtures, solutions, suspensions, colloids; the Tyndall effect; concentration calculations; and separation techniques with 20+ exam-style problems.

Atoms and Molecules

The atom is the indivisible building block — except when it isn't. Class 9 Atoms and Molecules with Dalton's atomic theory, laws of chemical combination, atomic and molecular masses, the mole concept, Avogadro's number, formula writing, and 25+ exam-style numericals with full step-by-step solutions.

Structure of the Atom

Dalton thought atoms were indivisible. He was wrong. Class 9 Structure of the Atom covers Thomson's plum-pudding model, Rutherford's gold-foil experiment, Bohr's planetary atom, the discovery of the neutron, electron-configuration rules, isotopes & isobars, valency and 25+ exam-style problems with full step-by-step solutions.