Describing Motion Around Us | CBSE Class 9 Science Notes

This note covers position and reference points, distance and displacement, average speed and velocity, average acceleration, position-time and velocity-time graphs, kinematic equations, stopping distances, and uniform circular motion.

How do we describe position, rest and motion?

Linear motion means motion along a straight line. A vertically falling ball, a car on a straight stretch of highway and a train on a straight track provide examples. Studying simplified forms of motion helps us understand more complicated movements.

What is a reference point?

A reference point is a fixed point chosen to describe an object's location. The object's position specifies its distance and direction from that point at a particular instant. Distance alone does not completely describe position: the direction from the reference point also matters.

Definition: An object is in motion when its position relative to a reference point changes with time. It is at rest when that position does not change with time.

On a straight line, mark the reference point as the origin, labelled O. Positions to its right are generally taken as positive and those to its left as negative. The signs + and − indicate the chosen directions along the line.

The origin and positive direction can be chosen for convenience. For a falling object, the release point can be the origin and downward can be positive. Once chosen, retain the same origin and positive direction throughout a calculation.

How does an instant differ from an interval?

An instant is a particular clock reading. A time interval is the duration between two clock readings. Position refers to an instant; a change in position compares positions at two instants. Confusing these ideas can lead to using the wrong time in a calculation.

Use m for metre, cm for centimetre and s for second when writing units. Context distinguishes a symbol for a quantity from a unit written after a number. For example, a time reading written with s after its numerical value is measured in seconds.

How do distance travelled and displacement differ?

Distance travelled measures the total length of the path followed. Displacement is the net change in position between two instants. Its direction is from the initial position towards the final position. Its magnitude is its numerical value with units, without specifying direction.

SI denotes the International System of Units. The SI unit of distance is the metre (m). The SI unit of displacement is the metre (m). Although their units match, distance and displacement describe different aspects of a journey.

The straight scale marks O at 0 m, B at 40 m and A at 100 m. The total-distance path goes from O to A and back to B; the displacement arrow runs directly from O to B.

See Fig. 4.4 in your NCERT textbook

What changes when the athlete turns back?

The athlete starts at O, reaches B after 4 s and A after 10 s, then returns to B after 16 s. The return journey adds to distance travelled, but the final displacement depends on the starting and stopping positions.

Worked example 1. An athlete runs from O at 0 m to A at 100 m, then back along the same track to B at 40 m. Find total distance and displacement, taking O towards A as positive.

Answer: Total distance = 100 m + 60 m = 160 m. Displacement = 40 m − 0 m = +40 m, towards B from O.

The magnitude of displacement is less than or equal to distance travelled. For straight-line motion, equality holds if the object moves without turning back. A ball moving vertically upwards and then downwards still follows a straight line, even though its direction reverses.

How are average speed and average velocity calculated?

Average speed is total distance travelled divided by the time interval. It describes how fast or slow the journey is overall and has no direction. Average velocity is displacement divided by the time interval, and its direction is the direction of displacement.

Let d represent total distance, s represent displacement, t represent the time interval, Vav represent average speed and vav represent average velocity. The capital and lower-case symbols keep the two averages distinct here.

Vav = d/t

vav = s/t

The SI unit of average speed is metre per second (m/s). The SI unit of average velocity is metre per second (m/s). The notation m s⁻¹ means the same as m/s. Kilometre per hour, written km/h or km h⁻¹, is another commonly used unit.

Why can average velocity be zero during a journey?

Worked example 2. Sarang swims to the opposite end of a 25 m pool and back to his starting point in 50 s. Find both averages.

Formula: Vav = d/t; vav = s/t.

Substitute: The complete path is 25 m + 25 m = 50 m, while displacement is 0 m.

Answer: Average speed is approximately 1 m/s, using 50 m ÷ 50 s. Average velocity is 0 m/s, using 0 m ÷ 50 s. Zero average velocity does not mean that Sarang remained at rest.

For straight-line motion in one direction, average speed equals the magnitude of average velocity. If the object returns to its starting position over a non-zero interval, its average velocity is zero even though its average speed is not zero.

What distinguishes uniform and non-uniform motion?

Uniform motion in a straight line covers equal distances in equal time intervals, for all possible choices of intervals. Its speed is constant. Non-uniform motion covers unequal distances in equal intervals; the speed increases, decreases or combines both changes.

A rate of change compares a change in a quantity with the corresponding time interval. Average velocity is the average rate of change of position. Velocity at a particular instant can differ from the average over a large interval.

What does acceleration tell us about motion?

Average acceleration is the change in velocity divided by the time interval. Define u as initial velocity, v as final velocity, t₁ as the initial time, t₂ as the final time and a as average acceleration over that interval.

a = (v − u)/(t₂ − t₁)

The SI unit of acceleration is metre per second squared (m/s²), also written m s⁻². Acceleration requires magnitude and direction. It can result from a change in velocity's magnitude, its direction, or both.

How is the direction of acceleration identified?

For straight-line motion, acceleration is in the direction of velocity when the magnitude of velocity increases. It is opposite to velocity when that magnitude decreases. Interpret signs using the chosen positive direction rather than treating a minus sign as a complete description by itself.

Worked example 3. A bus travelling along a straight highway increases its velocity from 36 km/h to 54 km/h in 10 s. Later, brakes bring it from 54 km/h to rest in 5 s. Take the forward direction as positive.

Answer: Convert the velocities to 10 m/s and 15 m/s. During speeding up, a = (15 − 10)/10 = +0.5 m/s², along the velocity. During braking, a = (0 − 15)/5 = −3 m/s², opposite to the velocity.

The conversion uses 1000 m per kilometre and 60 × 60 s per hour. Convert velocities before combining them with intervals measured in seconds. Keep the initial and final values matched to the particular interval being considered.

When is acceleration constant?

For straight-line motion in the same direction, acceleration is constant if velocity magnitude increases or decreases by equal amounts in equal time intervals, for all possible choices of intervals. With constant acceleration, the acceleration at each instant equals the average over any interval.

A falling object's successive downward velocities of 0, 9.8, 19.6, 29.4 and 39.2 m/s at 0, 1, 2, 3 and 4 s increase by 9.8 m/s each second. Its acceleration is 9.8 m/s² downward, denoted g, the acceleration due to Earth's gravitational force.

Note: High speed does not imply high acceleration. An object moving very fast can have zero acceleration if its velocity does not change. A vehicle's speedometer reading is nearly, but not exactly, the magnitude of its velocity at an instant.

How do we plot a position-time graph?

A position-time graph represents position against time. The horizontal X-axis shows time and the vertical Y-axis shows position. The axes meet at the graph's origin, O. A scale states how much of a quantity each graph-paper division represents.

The following positions describe a vehicle on a straight road. Read each time and position together as one pair. The zero-time position belongs at the origin because both values are zero.

What steps make the graph readable?

1. Draw perpendicular axes on graph paper and label the horizontal axis time and the vertical axis position.
2. Choose scales that fit the data conveniently: 5 divisions for 1 s horizontally and 5 divisions for 20 m vertically.
3. Mark values and units on both axes, retaining the chosen scale throughout each axis.
4. For each pair, locate its time and position. Their corresponding vertical and horizontal guide lines meet at the required point.
5. Plot every pair and join the points. These data form a straight line through the origin.

The intermediate points represent possible positions at intermediate times. They are correct if the vehicle moves with constant speed. Choosing a scale helps display data; it does not change the actual values represented by the graph.

Note: A motion graph is not a route map. A curved position-time graph describes how position changes with time; it does not show that the road itself curves.

The graph methods here concern straight-line motion in one direction. Distance travelled then equals displacement magnitude. If position is zero at time zero, the position-time graph also serves as the distance-time graph in this setting.

How do position-time graphs reveal velocity?

The slope of a line describes its steepness: change in the vertical-axis quantity divided by change in the horizontal-axis quantity. On a position-time graph, that ratio is change in position divided by elapsed time, which gives average velocity.

Let s₁ and s₂ denote the positions at times t₁ and t₂. For two points on the graph, use the difference between positions and the difference between times. Do not divide the final position by final time unless the chosen initial values make that appropriate.

vav = (s₂ − s₁)/(t₂ − t₁)

Worked example 4. A straight position-time graph gives position 40 m at 2 s and 80 m at 4 s. Find the velocity.

Answer: Change in position = 80 m − 40 m = 40 m. Time interval = 4 s − 2 s = 2 s. Velocity = 40 m ÷ 2 s = 20 m/s in the positive direction.

How do straight and curved graphs differ?

Panel (a) shows a rising straight position-time line, reaching 120 m at 6 s. Panel (b) shows an upward-curving line reaching 36 m at 12 s. Dashed guides compare changes in position over equal time intervals.

See Fig. 4.13 in your NCERT textbook

A rising straight line indicates constant velocity. The upward-curving example shows increasing displacement magnitudes in equal intervals, so the velocity magnitude increases. A horizontal position-time line indicates rest because position remains unchanged as time passes.

For example, a horizontal line at 40 m means the vehicle remains 40 m from the origin. It does not mean the vehicle travels 40 m repeatedly. This graph is not a distance-time graph for a journey starting at the origin.

When comparing straight position-time lines on the same axes, the steeper line represents greater velocity magnitude. For the same time interval, its change in position is larger. Compare slopes, rather than simply comparing how high two points appear.

What do the slope and area of a velocity-time graph show?

A velocity-time graph places time on the horizontal axis and velocity on the vertical axis. It shows whether velocity stays constant or changes. Its slope measures acceleration, while the area between the graph and time axis gives displacement over the chosen interval.

How does the line's shape describe acceleration?

A horizontal line at a non-zero velocity represents constant-velocity motion, with zero acceleration. A straight rising line represents constant acceleration along the velocity in the positive-motion examples. A straight falling line represents constant acceleration opposite to velocity while the object slows down.

The following two data sets show equal changes of velocity in equal time intervals. Each column can be plotted against time. Keep the decimal forms shown when transferring the values to a graph.

Between 10 s and 20 s, the rising line has acceleration (10 − 5)/(20 − 10) = 0.5 m/s². Over the same interval, the falling line gives −0.5 m/s². The sign distinguishes their acceleration directions for the chosen positive direction.

How does area give displacement?

At constant velocity, the area is a rectangle whose height is velocity and width is time interval. Their product has the unit metre and gives displacement. For a sloping straight line, split the area into a rectangle and triangle.

Worked example 5. A car moves at constant velocity 20 m/s for 6 s. Find displacement from its velocity-time graph.

Answer: The rectangular area gives displacement = 20 m/s × 6 s = 120 m, in the direction of motion.

The rising velocity-time line passes through A at 10 s and 5 m/s and B at 20 s and 10 m/s. The area ABDE is divided into rectangle ACDE and triangle ABC.

See Fig. 4.18(b) in your NCERT textbook

Worked example 6. A car's velocity rises uniformly from 5 m/s at 10 s to 10 m/s at 20 s. Find acceleration and displacement in this interval.

Formula: a = (v − u)/t; s = ut + ½(v − u)t. Here t is the interval of 10 s, rather than the final clock reading of 20 s.

Substitute: Acceleration = (10 − 5)/10. The rectangle contributes 5 × 10 and the triangle contributes ½ × 5 × 10.

Answer: Acceleration = 0.5 m/s². Displacement = 50 m + 25 m = 75 m, in the positive direction.

How are the equations for constant acceleration derived?

Kinematic equations relate displacement, time, initial velocity, final velocity and acceleration. They apply to straight-line motion with constant acceleration. Here u is velocity at time zero, v is velocity after interval t, a is the constant acceleration and s is displacement during that interval.

Derivation: Velocity after a given time

1. Constant acceleration equals average acceleration over the chosen interval, so a = (v − u)/t.
2. Multiply both sides by t to obtain at = v − u.
3. Add u to both sides to express the final velocity in terms of initial velocity, acceleration and time.

v = u + at

The product at is the change in velocity. This equation lets us predict velocity at a later time when initial velocity and constant acceleration are known. A negative acceleration is retained with its sign during substitution.

Derivation: Displacement from the graph

1. Draw the straight velocity-time line from initial velocity u to final velocity v over interval t. The enclosed area gives displacement.
2. The rectangular part has area ut. The triangular part has base t and height v − u, so its area is ½t(v − u).
3. Add these parts to obtain s = ut + ½t(v − u).
4. Use v − u = at and simplify the triangular contribution to ½at².

s = ut + ½at²

The velocity-time line rises from A at initial velocity u to B at final velocity v. The time interval is marked t. Rectangle OACD and triangle ABC together form the area OABD below the line.

See Fig. 4.19 in your NCERT textbook

How can time be eliminated?

Rearrange the velocity equation to obtain t = (v − u)/a for non-zero acceleration. Substitute this into the displacement equation. Combining the terms gives 2as = v² − u², so the equation without time is:

v² = u² + 2as

Select an equation according to the known quantities. Use v = u + at when displacement is unnecessary; use s = ut + ½at² when final velocity is unnecessary; use v² = u² + 2as when time is unnecessary.

The same constant-acceleration equations can also be rearranged into the following useful forms. Their conditions do not change just because their algebraic appearance changes.

s = ½(u + v)t

s = vt − ½at²

Note: These kinematic equations are valid only when acceleration is constant. If straight-line motion reverses direction, the signs of u, v, a and s specify directions. Displacement magnitude then need not equal the total distance travelled.

How do kinematic equations explain braking distance?

Braking distance is the distance covered while the vehicle slows to rest after the brakes act. A vehicle does not stop immediately when brakes are applied. The initial velocity and the acceleration caused by braking must both be considered.

How does initial velocity affect the calculation?

Worked example 7. A car moving in a straight line brakes with constant acceleration −4 m/s². Find its distance before stopping when its initial velocity is (i) 54 km/h and (ii) 108 km/h. Take forward as positive.

Answer: Final velocity v = 0 m/s. The initial velocities are 15 m/s and 30 m/s. From v² = u² + 2as, stopping distance s = (0 − u²)/(2 × (−4)). The distances are (i) 28.1 m and (ii) 112.5 m.

The initial velocity appears squared in this calculation. The larger initial velocity therefore produces a much greater distance before rest even though braking acceleration is unchanged. Retain the negative acceleration during substitution; the forward displacement before stopping is positive.

Why must a safe gap account for more than braking?

Reaction time is the interval before the driver responds by pressing the brake. Distance covered during this interval contributes to how far the vehicle travels before stopping. The road surface, braking capacity and initial velocity also affect the distance required.

A safe separation from the vehicle ahead must therefore allow for continued movement before rest. The calculation above isolates motion under the stated constant braking acceleration. A complete stopping problem can also require a separate calculation for movement during reaction time.

For a problem containing successive stages, identify where one stage ends and the next begins. Use the final velocity of one stage as the initial velocity of the next when the motion continues. Add the distances of the stages to find the total journey length.

Why is uniform circular motion accelerated?

Motion in a plane, or motion in two dimensions, includes a vehicle overtaking another, a kicked ball's path and a satellite moving on a circular path. Circular motion means motion along a circle. Uniform circular motion is circular motion at constant speed.

How do distance and displacement compare on a circle?

A revolution is one complete round. The radius is the distance from the circle's centre to its boundary, and the circumference is its boundary length. Write R for radius, T for time taken for one revolution and π for the circle constant, circumference divided by diameter. The diameter is the distance across the circle through its centre.

The top view of a merry-go-round marks A, B and C on a circle. The curved path passes from A through B to C; a straight dashed line joins A to C to represent displacement.

See Fig. 4.22 in your NCERT textbook

After one revolution, distance travelled is 2πR, but displacement is zero because the object returns to its starting position. Average speed is circumference divided by the revolution time; average velocity over that complete revolution is zero.

Vav = 2πR/T

In uniform circular motion, this speed is the same at every point on the circle. The zero average velocity over a complete revolution does not imply zero velocity at the individual instants during the motion.

Why does constant speed not imply constant velocity?

Velocity includes direction. On a circular path, its direction changes continuously even when its magnitude remains constant. A change in velocity means acceleration, so uniform circular motion is accelerated motion. No increase or decrease in speed is required for this conclusion.

A tangent is a straight line meeting the circle at one point. Velocity at a point on the circle lies along the tangent in the direction of motion. When a marble moving inside a ring is released by lifting the ring, it moves straight in its direction at release.

The conditions for uniform circular motion, constant speed and a circular path, are often not met in real-world situations. It is an idealised model that helps build an understanding of more complex motions, including planets revolving around the Sun and vehicles making circular turns.

Glossary

• Reference point — A fixed point relative to which an object's position is described using distance and direction.
• Position — The distance and direction of an object from a reference point at a particular instant.
• Distance travelled — The total length of the path followed by an object during a journey.
• Displacement — The net change in position between two instants, specified with magnitude and direction.
• Magnitude — The numerical value, with units, of a physical quantity, without stating its direction.
• Time interval — The duration between two instants, found from the difference between their clock readings.
• Average speed — Total distance travelled divided by the time interval during which that distance is covered.
• Average velocity — Displacement divided by the corresponding time interval, with direction matching that of displacement.
• Average acceleration — Change in velocity divided by the time interval over which the change occurs.
• Uniform linear motion — Straight-line motion covering equal distances in equal time intervals for all possible choices of intervals.
• Slope — A measure of line steepness obtained by dividing vertical change by corresponding horizontal change.
• Kinematic equations — Equations relating displacement, time, velocities and acceleration for straight-line motion with constant acceleration.
• Uniform circular motion — Motion along a circular path at constant speed, with continuously changing velocity direction.
• Tangent — A straight line that meets a circle at one and only one point.

Common errors and misconceptions

• Misconception: Distance travelled and displacement are interchangeable because both use metres. Correct: Distance measures the whole path; displacement measures net position change and includes direction. Turning back can make their magnitudes different.
• Misconception: Zero average velocity proves that an object remained at rest. Correct: An object can move and return to its starting point, giving zero displacement and therefore zero average velocity.
• Misconception: Equal distances in a few selected equal intervals prove uniform motion. Correct: Uniform straight-line motion requires equal distances in equal intervals for all possible choices of time intervals.
• Misconception: Fast motion necessarily means large acceleration. Correct: Acceleration concerns change in velocity. Even a very fast object has zero acceleration when its velocity remains constant.
• Misconception: A horizontal line means rest on every motion graph. Correct: It means rest on a position-time graph; at non-zero height on a velocity-time graph, it means constant non-zero velocity.
• Misconception: A curved position-time graph shows a curved road. Correct: It shows changing velocity in the straight-line examples. The graph represents position against time and is not a route map.
• Misconception: Kinematic equations can be applied regardless of how acceleration varies. Correct: The equations derived here require constant acceleration; check that condition before substituting values.
• Misconception: Constant speed means no acceleration in circular motion. Correct: The direction of velocity changes continuously around the circle, so uniform circular motion is accelerated despite its constant speed.

Exam-style questions with model answers

Q1. Explain how a reference point is used to distinguish rest from motion. [2 marks]

1. Choose a fixed reference point and describe the object's position by its distance and direction from that point.
2. If that position changes with time, the object is in motion; if it does not change, the object is at rest relative to that reference point.

Q2. Sarang swims along a 25 m pool to the opposite end and back to his starting point in 50 s. Find his total distance and displacement, average speed, and average velocity. [3 marks]

1. The outward path is 25 m and the return path is 25 m, so total distance is 50 m. Displacement is 0 m because the final and initial positions coincide.
2. Average speed is total distance divided by time: 50 m ÷ 50 s, approximately 1 m/s.
3. Average velocity is displacement divided by time: 0 m ÷ 50 s = 0 m/s.

Q3. A bus moving forward on a straight highway increases velocity from 36 km/h to 54 km/h in 10 s. Later it brakes from 54 km/h to rest in 5 s. Take forward as positive. Find both average accelerations and interpret their directions. Use 1 km = 1000 m and 1 h = 3600 s. [3 marks]

1. Convert the velocities to consistent units: 36 × 1000/3600 = 10 m/s and 54 × 1000/3600 = 15 m/s.
2. During speeding up, average acceleration = (15 − 10)/10 = +0.5 m/s². Its direction is forward, along the bus's velocity.
3. During braking, average acceleration = (0 − 15)/5 = −3 m/s². Its direction is opposite to the forward velocity, reducing the velocity magnitude.

Q4. A car's straight velocity-time graph rises from 5 m/s at 10 s to 10 m/s at 20 s. Motion is in the positive direction. Calculate the interval, acceleration, rectangular area contribution and total displacement between these instants. [4 marks]

1. The required time interval is 20 s − 10 s = 10 s. Use this interval rather than the final clock reading.
2. The slope gives acceleration: (10 − 5)/10 = 0.5 m/s² in the positive direction.
3. The rectangular contribution to area is initial velocity × interval = 5 × 10 = 50 m.
4. The triangular contribution is ½ × 10 × (10 − 5) = 25 m. Total displacement is 50 + 25 = 75 m in the positive direction.

Q5. Derive s = ut + ½at² using a velocity-time graph for straight-line motion with constant acceleration. Here s is displacement, u initial velocity, v final velocity, a acceleration and t elapsed time. Consider motion with increasing positive velocity. [5 marks]

1. Draw time on the horizontal axis and velocity on the vertical axis. Constant acceleration gives a straight line from initial velocity u at time zero to final velocity v at time t.
2. The area between the line and time axis over this interval represents displacement. Divide it into a rectangle and a triangle.
3. The rectangle has height u and width t, giving area ut. The triangle has height v − u and base t, giving area ½t(v − u).
4. Add the two contributions: s = ut + ½t(v − u). From the definition of constant acceleration, a = (v − u)/t, so v − u = at.
5. Substitute at for v − u: s = ut + ½t(at) = ut + ½at². This result requires constant acceleration throughout the interval.

Q6. A car brakes with constant acceleration −4 m/s² while moving in a straight line. Compare stopping distances from initial velocities 54 km/h and 108 km/h. Take forward as positive and use 1 km = 1000 m and 1 h = 3600 s. State the final velocity, convert the initial velocities, select the equation and calculate each distance. [5 marks]

1. The car stops, so its final velocity is 0 m/s in both cases. The acceleration during braking is the given constant value −4 m/s².
2. Convert the initial velocities before calculating: 54 × 1000/3600 = 15 m/s, and 108 × 1000/3600 = 30 m/s.
3. Use v² = u² + 2as, with u initial velocity, v final velocity, a acceleration and s displacement. Rearranging gives s = (v² − u²)/(2a).
4. For 15 m/s, s = (0 − 15²)/(2 × (−4)), giving 28.1 m to one decimal place. Distance equals displacement magnitude before stopping.
5. For 30 m/s, s = (0 − 30²)/(2 × (−4)) = 112.5 m. The higher initial velocity requires a much greater distance with the same braking acceleration.

Q7. Explain why uniform circular motion is accelerated, and derive its speed expression for radius R and revolution time T. State the average velocity over one full revolution. [3 marks]

1. The speed stays constant, but the direction of velocity changes continuously. Since velocity changes, the motion is accelerated.
2. One revolution covers circumference 2πR, where π is the circle constant. Dividing this distance by time T gives speed 2πR/T.
3. After a complete revolution, the object returns to its starting position. Displacement is zero, so average velocity over that revolution is zero despite the non-zero speed.

Q8. Compare a horizontal position-time graph at 40 m with a horizontal velocity-time graph at 20 m/s. Explain the motion in each case and state the acceleration in the second case. [3 marks]

1. The horizontal position-time line means position stays at 40 m while time changes. The object is at rest 40 m from the origin.
2. The horizontal velocity-time line means the object keeps moving with constant velocity 20 m/s. Its position changes as time passes.
3. The second object's acceleration is zero because its velocity does not change. The graph's slope is zero; its non-zero height represents velocity, not acceleration.

Key takeaways

• Position requires both distance and direction from a reference point; changes in that position distinguish motion from rest.
• Distance counts the whole path, while displacement records net position change and can be zero after a journey.
• Average speed uses total distance divided by time; average velocity uses displacement divided by the same time interval.
• Acceleration measures change in velocity, which can involve a change in magnitude, direction, or both together.
• A position-time slope gives average velocity between two points; a velocity-time slope gives average acceleration between two points.
• The area between a velocity-time graph and the time axis gives displacement during the selected time interval.
• The kinematic equations require constant acceleration; retain the signs of directional quantities and use consistent units in substitutions.
• Uniform circular motion has constant speed and changing velocity direction, making it accelerated motion with zero displacement after each revolution.

Test yourself

What information is needed to specify position?

Specify the object's distance and direction from a chosen reference point at a particular instant.

When do distance travelled and displacement magnitude match in straight-line motion?

They match when the object moves in one direction without turning back during the interval considered.

How can a moving object have zero average velocity over an interval?

It can return to its initial position. Its displacement is then zero, so displacement divided by elapsed time is zero.

Does a minus sign on acceleration by itself show whether speed increases?

No. Compare acceleration with velocity direction. Speed increases when acceleration is along velocity and decreases when it is opposite.

What does a rising straight position-time line indicate?

It indicates constant velocity: equal position changes occur in equal time intervals, so the line has a constant slope.

What does area under a velocity-time graph represent?

For the straight-line, one-direction cases considered here, area between the graph and time axis gives displacement over the chosen interval.

What condition must hold when using v² = u² + 2as?

Acceleration must be constant for the straight-line motion. Here u and v are initial and final velocities, a acceleration and s displacement.

Why is uniform circular motion an idealised model?

Its conditions, constant speed and a circular path, are often not met in real-world situations, though the model helps explain more complex motions.