Motion — Class 9 Science Notes (NCERT)

Snapshot

Scalars have magnitude only (distance, speed, time, mass). Vectors have magnitude AND direction (displacement, velocity, acceleration, force).
Three equations of motion (uniform acceleration only): v = u + at | s = ut + (1/2)at² | v² = u² + 2as.
On a distance-time graph, slope = speed. On a velocity-time graph, slope = acceleration and area under curve = distance covered.
Uniform circular motion: speed is constant but velocity changes direction every instant, so there is acceleration (centripetal) even though speed is unchanged.
Board weightage: ~8 marks/year — typically one 3-mark numerical (equations of motion), one 2-mark graph question, one 3-mark derivation or conceptual short answer.

Detailed Notes

1. Rest and Motion

An object is said to be in rest when its position does not change with time relative to its surroundings. An object is said to be in motion when its position changes with time relative to its surroundings.

Key idea: rest and motion are relative. A passenger sitting in a moving bus is at rest relative to fellow passengers, but is in motion relative to a person standing on the road. There is no absolute rest in nature — every object on Earth is moving with Earth's rotation and revolution around the Sun.

Reference point (origin): To describe the position or motion of an object, we always choose a fixed reference point. The position is then measured as the distance from the reference point in a specific direction.

Types of motion by path shape:

Rectilinear (linear) motion — along a straight line, e.g., a car on a straight road, a stone dropped vertically.
Circular motion — along a circular path, e.g., a satellite orbiting Earth, a stone tied to string whirled horizontally.
Oscillatory or vibratory motion — to and fro about a fixed point, e.g., a pendulum, a vibrating guitar string.

In this chapter, NCERT focuses mainly on rectilinear (straight-line) motion under uniform acceleration, and closes with uniform circular motion.

2. Distance vs. Displacement

These two quantities are the most confused pair in this chapter. The core difference is direction.

Distance	Displacement
Total path length actually covered by the object	Shortest straight-line distance from starting point to ending point, with direction
Scalar (magnitude only)	Vector (magnitude + direction)
Always positive or zero; never negative	Can be positive, negative, or zero
Never decreases; keeps accumulating	Can decrease if the object retraces its path
SI unit: metre (m)	SI unit: metre (m)

Classic example: A man walks 4 km East and then 3 km North.

Distance = 4 + 3 = 7 km
Displacement = shortest path from start to end = √(4² + 3²) = √25 = 5 km, in the direction North-East

Important special cases:

If an object moves in a straight line in one direction only: distance = magnitude of displacement.
If an object returns to its starting point: displacement = 0, but distance > 0.
For one complete lap on a circular track: displacement = 0, distance = circumference of track.
Golden rule: Distance ≥ |Displacement|, always.

3. Speed vs. Velocity — Scalar vs. Vector

Speed is the rate of change of distance with time. Velocity is the rate of change of displacement with time.

$$\text{Speed} = \frac{\text{Distance covered}}{\text{Time taken}} \qquad\qquad \text{Velocity} = \frac{\text{Displacement}}{\text{Time taken}}$$

Speed	Velocity
Scalar quantity	Vector quantity
Always ≥ 0; never negative	Can be negative (motion in opposite direction)
Speed = distance / time	Velocity = displacement / time
SI unit: m/s or m s¹	SI unit: m/s or m s¹ (with direction)

Average speed vs. average velocity:

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \qquad\qquad \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$$

Uniform speed: equal distances covered in equal time intervals. Non-uniform speed: unequal distances in equal time intervals (e.g., city traffic).

Odometer records total distance; speedometer gives instantaneous speed.

Unit conversion (very important for numericals):

$$1 \text{ km/h} = \frac{5}{18} \text{ m/s} \qquad\qquad 1 \text{ m/s} = 3.6 \text{ km/h}$$

4. Uniform vs. Non-Uniform Motion

Uniform motion: An object moving along a straight line covers equal distances in equal intervals of time, however small the time intervals may be. The speed is constant throughout. Example: a car travelling on cruise control on a straight highway at 80 km/h.

Non-uniform motion: An object covers unequal distances in equal intervals of time. The speed keeps changing. Examples: a bicycle in city traffic, a freely falling stone (speed increases every second), a car braking at a signal.

How to tell from a graph:

Straight slanted line on distance-time graph ⇒ uniform motion.
Curved line on distance-time graph ⇒ non-uniform motion (changing speed).
Horizontal line on velocity-time graph ⇒ uniform motion (constant velocity).
Slanted straight line on velocity-time graph ⇒ uniformly accelerated motion.

Instantaneous speed: The speed at a specific instant. As the time interval approaches zero, instantaneous speed = rate of change of distance at that point. For uniform motion, instantaneous speed = average speed = constant.

5. Acceleration — Formula, Units, Sign

Acceleration is the rate of change of velocity with time. It measures how quickly velocity changes.

$$a = \frac{v - u}{t} = \frac{\text{Change in velocity}}{\text{Time taken}}$$

where u = initial velocity, v = final velocity, t = time taken.

SI unit: m s² (metres per second per second) or m/s²
Vector quantity — has both magnitude and direction.
Positive acceleration: velocity increases in the direction of motion. Example: car speeding up from 0 to 60 km/h.
Negative acceleration (retardation / deceleration): velocity decreases; acceleration is opposite to direction of motion. Example: brakes applied. Written as a negative value in equations.
Zero acceleration: uniform motion — velocity does not change at all.

Uniform acceleration: velocity changes by equal amounts in equal time intervals. Examples: free fall under gravity (g = 9.8 m/s² ≈ 10 m/s²), a ball rolling down a smooth inclined plane.

Non-uniform acceleration: velocity changes by unequal amounts in equal time intervals. Example: a car in city traffic.

Important note: An object can be accelerating even while its speed stays constant — if its direction changes (e.g., uniform circular motion). Acceleration requires a change in velocity, and velocity changes if either speed or direction changes.

6. Distance-Time Graph — Slope = Speed

A distance-time (s-t) graph has distance (in metres) on the y-axis and time (in seconds) on the x-axis. The slope of this graph gives the speed of the object.

$$\text{Slope of distance-time graph} = \frac{\Delta s}{\Delta t} = \textbf{Speed}$$

Reading a distance-time graph:

Horizontal line (slope = 0): object is at rest. Distance is not changing with time.
Straight line with positive slope: uniform motion (constant speed). A steeper line means higher speed.
Curve bending upward (concave up): non-uniform motion — speed is increasing (object accelerating).
Curve bending downward (concave down): non-uniform motion — speed is decreasing (object decelerating).

How to calculate speed from graph: Pick two points P1 = (t1, d1) and P2 = (t2, d2) on the straight line.

$$\text{Speed} = \frac{d_2 - d_1}{t_2 - t_1}$$

Key limitation: On a pure distance-time graph, the line can never have a negative slope (distance cannot decrease). On a displacement-time graph, negative slope means the object is moving back towards the origin.

7. Velocity-Time Graph — Slope = Acceleration, Area = Distance

A velocity-time (v-t) graph plots velocity (y-axis) against time (x-axis). This is the most important graph in the chapter — both equations of motion and distances are derived from it.

$$\text{Slope of v-t graph} = \frac{\Delta v}{\Delta t} = \textbf{Acceleration} \qquad\qquad \text{Area under v-t graph} = \textbf{Distance covered}$$

Reading a velocity-time graph:

Horizontal straight line (slope = 0): uniform velocity, zero acceleration. Area = rectangle = v × t = distance.
Straight line with positive slope: uniform positive acceleration (velocity increasing). Area = trapezium.
Straight line with negative slope: uniform retardation (velocity decreasing towards zero).
Curved line: non-uniform acceleration (changing rate of velocity change).

Why does area = distance? In a very small time interval Δt, the distance covered = v × Δt = a thin strip of area under the graph. Adding up all the tiny strips from start to finish gives the total area = total distance covered. This is a universal result, valid for any shape of v-t graph.

Area formulas to remember:

For uniform velocity (horizontal line): Area = rectangle = v × t
For uniform acceleration from u to v: Area = trapezium = ½(u + v) × t
Split into: rectangle (u × t) + triangle (½(v−u) × t) = ut + ½at²

8. Equations of Motion (for Uniform Acceleration)

When an object moves with uniform (constant) acceleration along a straight line, three equations relate the five quantities: u (initial velocity), v (final velocity), a (acceleration), t (time), s (distance covered).

First Equation of Motion: v = u + at

$$\boxed{v = u + at}$$

Algebraic derivation (from definition of acceleration):

By definition: a = (v − u) / t ⇒ at = v − u ⇒ v = u + at

Use this equation when: you know u, a, t and want to find v. (This equation does not contain 's'.)

Second Equation of Motion: s = ut + (1/2)at²

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

Algebraic derivation (from average velocity):

Average velocity = (u + v) / 2. Also, distance s = average velocity × time:

s = [(u + v) / 2] × t. Substitute v = u + at:

s = [(u + u + at) / 2] × t = [(2u + at) / 2] × t = ut + ½at²

Use this equation when: you know u, a, t and want to find s. (This equation does not contain 'v'.)

Third Equation of Motion: v² = u² + 2as

$$\boxed{v^2 = u^2 + 2as}$$

Algebraic derivation (eliminate t):

From 1st equation: t = (v − u) / a. Substitute in s = ½(u + v)t:

s = ½(u + v) × (v − u) / a = (v² − u²) / (2a)

Therefore: 2as = v² − u² ⇒ v² = u² + 2as

Use this equation when: time 't' is not given or not needed. (This equation does not contain 't'.)

Quick selector — which equation to use:
v = u + at → no s (use when distance is not involved)
s = ut + ½at² → no v (use when final velocity is not involved)
v² = u² + 2as → no t (use when time is not involved)

Critical condition: All three equations are valid ONLY for uniform (constant) acceleration along a straight line. They do NOT apply to circular motion or non-uniform acceleration.

Special values to remember: If object starts from rest: u = 0. If object comes to rest: v = 0. Free fall: a = g = 9.8 m/s² (downward); for upward throw: a = −g = −9.8 m/s².

9. Uniform Circular Motion

When an object moves along a circular path with constant speed, it is said to be in uniform circular motion.

$$\text{Speed} = \frac{\text{Circumference of circle}}{\text{Time period}} = \frac{2\pi r}{T}$$

where r = radius of the circular path, T = time period (time for one complete revolution).

Why is it accelerated motion?

Speed is constant, but the direction of motion changes at every point on the circle. Since velocity = speed + direction, and direction keeps changing, the velocity keeps changing. Since acceleration = rate of change of velocity, there IS acceleration even though speed is constant. This acceleration always points toward the centre of the circle and is called centripetal acceleration. The force producing it is called centripetal force.

Key points about uniform circular motion:

Speed = constant throughout; velocity = not constant (direction changes continuously).
Velocity is always directed tangentially to the circle at any point.
Acceleration (centripetal) points toward the centre of the circle at every point.
The three equations of motion (v = u + at etc.) do NOT apply here.
One complete revolution: displacement = 0, but distance = 2πr.

Examples of uniform circular motion:

A stone tied to a string and whirled in a horizontal circle at constant speed.
The Moon revolving around the Earth (approximately circular orbit).
An artificial satellite in a circular orbit around Earth.
The tip of the second hand of a clock tracing a circle at constant speed.
An athlete running at constant speed on a circular track.
A cyclist going around a circular track at constant speed.

10. Graphical Method of Deriving Equations of Motion

NCERT derives all three equations of motion using a velocity-time graph. This is a standard 3-mark question. Always draw and label the graph carefully.

Setup: Consider an object with initial velocity OA = u at time t = 0, accelerating uniformly to final velocity BC = v at time OC = t. On the v-t graph, plot point A at (0, u) and point B at (t, v). Line AB is straight (uniform acceleration).

Draw BC perpendicular to the time axis (down to C on x-axis). Draw AD parallel to the time axis (perpendicular from A, meeting BC at D). Now: OA = DC = u, BD = BC − DC = v − u, OC = AD = t.

Graphical derivation — Equation 1 (v = u + at)

Slope of line AB = acceleration a = (BD) / (AD) = (v − u) / t

So: at = v − u ⇒ v = u + at (Proved)

Graphical derivation — Equation 2 (s = ut + (1/2)at²)

Distance covered s = area of trapezium OABC = area of rectangle OADC + area of triangle ABD

= (OA × OC) + (½ × AD × BD)

= (u × t) + (½ × t × (v − u))

= ut + ½ × t × at (since v − u = at)

= ut + ½at² (Proved)

Graphical derivation — Equation 3 (v² = u² + 2as)

Distance s = area of trapezium OABC = ½ × (OA + BC) × OC = ½(u + v) × t

From 1st equation: t = (v − u) / a. Substituting:

s = ½(u + v) × (v − u) / a = (v² − u²) / (2a)

So: 2as = v² − u² ⇒ v² = u² + 2as (Proved)

NCERT Worked Examples — Fully Solved

NCERT Example 8.1 — Average speed and average velocity (straight-line motion)

An object travels 16 m in 4 s and then another 16 m in 2 s. Find average speed and average velocity. (Motion is along a straight line in the same direction throughout.)

Given: First stretch: s1 = 16 m, t1 = 4 s. Second stretch: s2 = 16 m, t2 = 2 s.

Total distance = 16 + 16 = 32 m | Total displacement = 32 m (same direction) | Total time = 4 + 2 = 6 s

Average speed = Total distance / Total time = 32 / 6 = 5.33 m/s

Average velocity = Total displacement / Total time = 32 / 6 = 5.33 m/s (same because motion is in one direction)

Note: For motion in one direction, average speed = magnitude of average velocity.

NCERT Example 8.2 — Speed at different parts, find average speed

Usman travels 300 km by road to visit his friend. For the first 200 km, he drives at 60 km/h. For the remaining 100 km, he drives at 40 km/h. Find his average speed for the entire journey.

Time for first 200 km: t1 = 200 / 60 = 10/3 h

Time for remaining 100 km: t2 = 100 / 40 = 5/2 h

Total time: t = 10/3 + 5/2 = 20/6 + 15/6 = 35/6 h

Total distance: = 300 km

Average speed = 300 / (35/6) = 300 × 6/35 = 1800/35 = 51.4 km/h

Key insight: Average speed is NOT (60 + 40)/2 = 50 km/h because equal distances (not equal times) are covered at each speed.

NCERT Example 8.3 — Bus attains velocity, find acceleration

A bus starting from rest attains a velocity of 54 km/h in 3 minutes. Find the acceleration and the distance covered by the bus.

Given: u = 0 (starts from rest), v = 54 km/h = 54 × 5/18 = 15 m/s, t = 3 min = 3 × 60 = 180 s.

(i) Acceleration: a = (v − u) / t = (15 − 0) / 180 = 15/180 = 1/12 m/s² = 0.083 m/s²

(ii) Distance: s = ut + ½at² = 0 + ½ × (1/12) × 180² = (1/24) × 32400 = 1350 m = 1.35 km

Check: v² = u² + 2as = 0 + 2 × (1/12) × 1350 = 2700/12 = 225 = 15². Confirmed.

NCERT Example 8.4 — Train reaches 72 km/h in 5 minutes

A train starting from a railway station and moving with uniform acceleration attains a speed of 72 km/h in 10 minutes. Find the acceleration.

Given: u = 0, v = 72 km/h = 72 × 5/18 = 20 m/s, t = 10 min = 10 × 60 = 600 s.

Acceleration: a = (v − u) / t = (20 − 0) / 600 = 20/600 = 1/30 m/s² = 0.033 m/s²

Distance covered: s = ut + ½at² = 0 + ½ × (1/30) × 600² = (1/60) × 360000 = 6000 m = 6 km

NCERT Example 8.5 — Retardation: car braking to stop

A car travelling at 90 km/h applies brakes and stops after covering 50 m. Find the retardation produced by the brakes.

Given: u = 90 km/h = 90 × 5/18 = 25 m/s, v = 0 (car stops), s = 50 m.

Using v² = u² + 2as: 0 = 25² + 2 × a × 50 = 625 + 100a

100a = −625 ⇒ a = −6.25 m/s²

Retardation = 6.25 m/s² (magnitude; the negative sign shows deceleration)

Time to stop: v = u + at ⇒ 0 = 25 + (−6.25)t ⇒ t = 25/6.25 = 4 s

NCERT Example 8.6 — Uniform circular motion: speed of satellite

An artificial satellite completes one orbit around the Earth in 90 minutes. If the radius of the Earth is 6400 km and the satellite orbits at a height of 800 km above the Earth's surface, find the orbital speed of the satellite.

Given: T = 90 min = 90 × 60 = 5400 s, Height h = 800 km, Radius of Earth R = 6400 km.

Radius of orbit: r = R + h = 6400 + 800 = 7200 km = 7,200,000 m = 7.2 × 10&sup6; m.

Speed: v = 2πr / T = (2 × 3.14 × 7.2 × 10&sup6;) / 5400

= (45,216,000) / 5400 = 8374 m/s ≈ 8.4 km/s

This enormous speed (≈ 8.4 km per second) keeps the satellite in orbit against Earth's gravity.

Extra NCERT-style — Object thrown vertically upward

A ball is thrown vertically upward with an initial velocity of 49 m/s. Find (i) the maximum height reached and (ii) the time taken to reach maximum height. (Take g = 9.8 m/s².)

Given: u = 49 m/s (upward, positive direction), v = 0 (at maximum height, momentarily stops), a = −9.8 m/s² (gravity acts downward, opposite to motion).

(i) Maximum height: Using v² = u² + 2as:

0 = 49² + 2 × (−9.8) × s = 2401 − 19.6s

19.6s = 2401 ⇒ s = 2401 / 19.6 = 122.5 m

(ii) Time to reach top: Using v = u + at:

0 = 49 + (−9.8) × t ⇒ 9.8t = 49 ⇒ t = 5 s

Extra NCERT-style — Stone dropped from tower, find velocity on impact

A stone is dropped from the top of a tower 80 m high. Find (i) the velocity with which the stone hits the ground and (ii) the time taken. (Take g = 10 m/s².)

Given: u = 0 (dropped, not thrown), s = 80 m (downward), a = +10 m/s² (gravity, same direction as motion).

(i) Velocity on impact: v² = u² + 2as = 0 + 2 × 10 × 80 = 1600 ⇒ v = √1600 = 40 m/s

(ii) Time taken: v = u + at ⇒ 40 = 0 + 10t ⇒ t = 4 s

Practice MCQs

1. An athlete runs once around a 400 m circular track and finishes at the starting point. Which statement is correct?

Distance = 0, Displacement = 400 m
Distance = 400 m, Displacement = 0
Both distance and displacement = 400 m
Both distance and displacement = 0

Answer: (B) Distance = 400 m, Displacement = 0. Distance = total path covered = one full lap = 400 m. Displacement = shortest path from start to end = 0 because the athlete finishes at the starting point. This is the classic example of distance ≠ displacement.

2. Which of the following is a vector quantity?

Distance
Speed
Mass
Velocity

Answer: (D) Velocity. Velocity has both magnitude (how fast) and direction (which way). Distance, speed, and mass are all scalar quantities with magnitude only.

3. A car starts from rest and accelerates uniformly at 3 m/s² for 6 s. What distance does it cover?

18 m
27 m
54 m
108 m

Answer: (C) 54 m. u = 0, a = 3 m/s², t = 6 s. Using s = ut + ½at² = 0 + ½ × 3 × 36 = 54 m.

4. On a distance-time graph, a straight horizontal line indicates:

Uniform motion
The object is at rest
Uniformly accelerated motion
Non-uniform motion

Answer: (B) The object is at rest. A horizontal line means distance does not change with time — the object is stationary. Slope = 0 = speed = 0.

5. The area under a velocity-time graph gives:

Acceleration of the object
Speed of the object
Distance covered by the object
Rate of change of acceleration

Answer: (C) Distance covered by the object. Area under v-t graph = velocity × time = distance. This result is fundamental and applies to any shape of v-t graph.

6. A stone is dropped from the top of a tower. It hits the ground in 4 s. (Take g = 10 m/s².) The height of the tower is:

20 m
40 m
80 m
160 m

Answer: (C) 80 m. u = 0 (dropped), a = 10 m/s², t = 4 s. s = ut + ½at² = 0 + ½ × 10 × 16 = 80 m.

7. A car moving at 20 m/s applies brakes and decelerates at 5 m/s². How long does it take to stop?

2 s
4 s
5 s
10 s

Answer: (B) 4 s. u = 20 m/s, v = 0, a = −5 m/s². Using v = u + at: 0 = 20 − 5t ⇒ t = 20/5 = 4 s.

8. In uniform circular motion, which of the following is TRUE?

Both speed and velocity are constant
Speed is constant but velocity changes continuously
Velocity is constant but speed changes
Both speed and velocity change continuously

Answer: (B) Speed is constant but velocity changes continuously. The magnitude of velocity (speed) stays the same. But direction changes at every point on the circle, so velocity (a vector) changes. Hence there is centripetal acceleration directed toward the centre.

9. Which equation of motion does NOT contain the variable 's' (distance/displacement)?

v = u + at
s = ut + ½at²
v² = u² + 2as
All three contain 's'

Answer: (A) v = u + at. This equation relates v, u, a, and t — 's' does not appear. The second equation lacks 'v'; the third equation lacks 't'. These absences tell you which equation to pick for a given problem.

10. A cyclist goes around a circular track of radius 70 m in 44 s at constant speed. (Take π = 22/7.) His speed is:

5 m/s
10 m/s
15 m/s
20 m/s

Answer: (B) 10 m/s. Circumference = 2πr = 2 × (22/7) × 70 = 2 × 22 × 10 = 440 m. Speed = 440/44 = 10 m/s.

Previous-Year Questions (CBSE)

PYQ 1. Define uniform acceleration. A car starts from rest and attains a velocity of 72 km/h in 10 s. Calculate the acceleration. (CBSE 2023, 3 marks)

Uniform acceleration: When the velocity of an object changes by equal amounts in equal intervals of time, the motion is called uniformly accelerated motion.

Given: u = 0 (starts from rest), v = 72 km/h = 72 × 5/18 = 20 m/s, t = 10 s.

a = (v − u) / t = (20 − 0) / 10 = 2 m/s²

PYQ 2. Derive the equation v = u + at using a velocity-time graph. (CBSE 2022, 3 marks)

Draw a v-t graph. Let initial velocity OA = u at t = 0. Let final velocity BC = v at time OC = t. Draw AD parallel to x-axis meeting BC at D. So DC = u, BD = v − u, AD = t.

Slope of AB = acceleration = BD / AD = (v − u) / t

Therefore: at = v − u ⇒ v = u + at (Proved)

PYQ 3. A stone is thrown vertically upward with initial velocity 5 m/s. (g = 10 m/s².) Find maximum height reached and total time of flight. (CBSE 2022, 3 marks)

Given: u = 5 m/s (upward), v = 0 (at maximum height), a = −10 m/s².

Max height: v² = u² + 2as ⇒ 0 = 25 − 20s ⇒ s = 25/20 = 1.25 m

Time to reach top: v = u + at ⇒ 0 = 5 − 10t ⇒ t = 0.5 s

Total time of flight = 2 × 0.5 = 1 s (time up = time down for symmetric projectile)

PYQ 4. Distinguish between speed and velocity. Give one example where speed is non-zero but velocity is zero. (CBSE 2021, 3 marks)

Speed: Rate of change of distance with time. Scalar. Always positive or zero.

Velocity: Rate of change of displacement with time. Vector. Can be positive, negative, or zero.

Example: A runner completes one full lap (400 m) around a circular track in 80 s. Speed = 400/80 = 5 m/s (non-zero). Displacement = 0 (returns to start). Velocity = 0/80 = 0 m/s (zero).

PYQ 5. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at 10 m/s², with what velocity does it strike the ground? What is its velocity 1 s after it is dropped? (CBSE 2020, 3 marks)

Given: u = 0 (gently dropped), s = 20 m, a = 10 m/s².

Velocity on striking ground: v² = u² + 2as = 0 + 2 × 10 × 20 = 400 ⇒ v = 20 m/s

Velocity after 1 s: v = u + at = 0 + 10 × 1 = 10 m/s

Common Mistakes and How to Avoid Them

Not converting km/h to m/s: Always multiply km/h by 5/18 before substituting in equations of motion. 36 km/h = 10 m/s; 72 km/h = 20 m/s; 90 km/h = 25 m/s.
Applying equations of motion to non-uniform acceleration: The three equations are ONLY valid for constant (uniform) acceleration in a straight line. Never use them for circular motion or varying acceleration.
Dropping the negative sign for retardation: When an object decelerates, a is negative. Write it as −5 m/s², not +5 m/s². Substituting the wrong sign gives wrong answers.
Confusing distance and displacement in word problems: When a body returns to the starting point, displacement = 0. Calculate displacement carefully using directions, not just adding distances.
Slope of distance-time graph vs. velocity-time graph: Slope of d-t = speed. Slope of v-t = acceleration. These are different — do not mix up the two graphs.
Circular motion misconception: Students often say "acceleration is zero because speed is constant in circular motion." WRONG. Acceleration exists because direction (and hence velocity) keeps changing. Correct answer: acceleration is centripetal (toward centre).
Omitting direction in displacement or velocity answers: These are vectors. A full answer must include direction (e.g., 5 m East, not just 5 m).

Quick Revision Checklist

Distance = total path covered (scalar); displacement = shortest straight-line path with direction (vector).
Speed = distance/time (scalar); velocity = displacement/time (vector).
Acceleration = (v − u)/t; SI unit = m/s²; negative value = retardation.
Distance-time graph: slope = speed; horizontal = rest; curve = changing speed.
Velocity-time graph: slope = acceleration; area under graph = distance; horizontal = uniform velocity.
v = u + at (no s); s = ut + ½at² (no v); v² = u² + 2as (no t).
All three equations valid ONLY for uniform acceleration along a straight line.
Uniform circular motion: constant speed, changing velocity ⇒ centripetal acceleration toward centre.
Speed in uniform circular motion: v = 2πr/T.
Conversion: 1 km/h = 5/18 m/s; 1 m/s = 3.6 km/h = 18/5 km/h.
Free fall: a = g = 9.8 m/s² downward; for upward throw: a = −g.
Distance ≥ |Displacement|; average speed ≥ |average velocity| always.

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