Motion in a Straight Line — Part 1

What is Motion?

Look around you. A fan spinning, a car driving down the road, a bird flying in the sky, even the tiny dust particles dancing in a sunbeam. Our world is in a constant state of flux. In science, we call this change in an object's position with time, motion.

But consider this: are you moving right now? You might say, "No, I'm sitting still." But what if someone was watching you from the Moon? To them, you, your chair, and your entire city are hurtling through space as the Earth rotates and revolves around the Sun.

This tells us something fundamental: motion is relative. An object's motion depends entirely on the observer's point of view. The person sitting next to you on a moving bus is at rest relative to you, but both of you are in motion relative to a person standing on the sidewalk.

To describe motion accurately and without confusion, we must first agree on a stationary background or a point of reference.

{{KEY: type=definition | title=Motion | text=An object is said to be in motion if its position changes continuously with respect to a stationary object, taken as a reference point.}}

The Reference Point (Origin)

Imagine you need to give a friend directions to your house. Saying "My house is 500 metres away" is not very helpful. 500 metres from where? In which direction?

A much better instruction would be, "From the big banyan tree at the crossroads, walk 500 metres north." Here, the "big banyan tree" is your reference point, also known as the origin. It's a fixed point in space from which we measure the position of other objects. By specifying a distance and a direction from this origin, you can describe any location precisely.

In physics, we often use a coordinate system, like a number line or a graph, with the origin marked as 0 to describe positions.

{{VISUAL: diagram: A straight horizontal line representing a road, marked with kilometers like a ruler. A tree is shown at the '0 km' mark, labelled 'Origin'. A red car is shown at the '+40 km' mark, and a blue bus is at the '-20 km' mark.}}

The Simplest Journey: Motion Along a Straight Line

The most basic type of motion to study is when an object moves along a straight path. This is called rectilinear motion (from rectus, Latin for 'straight', and linea, for 'line'). Think of a train on a straight track, a sprinter running the 100-metre dash, or an apple falling from a tree.

Let's use our straight-line road from the diagram above to understand two crucial terms we use to describe a journey: distance and displacement. These words might seem similar in everyday language, but in physics, they have very different meanings.

Distance and Displacement: Two Ways to Measure a Journey

Imagine our red car starts its journey from the origin O (the tree at 0 km).

1. It first travels to a town A located at +60 km.
2. Then, it turns around and travels back to a town B located at +35 km.

How can we describe this entire journey? We have two ways.

The Total Path: Distance

The most straightforward way is to measure the total path length the car has covered.

• From O to A, the car travelled 60 km.
• From A back to B, the car travelled from the 60 km mark to the 35 km mark. The length of this path is 60 km - 35 km = 25 km.

The total distance covered by the car is the sum of these two parts: Distance = 60 km + 25 km = 85 km

Distance only cares about the total ground covered, regardless of the direction of travel. Because it only has a numerical value (magnitude) and no associated direction, it is a scalar quantity.

{{KEY: type=definition | title=Distance | text=Distance is the total length of the path covered by a moving object from its initial position to its final position. Its SI unit is the metre (m).}}

The Shortest Path: Displacement

Now, let's look at the journey differently. Where did the car start? At O (position 0 km). Where did it end? At B (position +35 km).

The displacement is the change in the object's position. It is the shortest, straight-line distance between the starting point and the ending point, measured with direction.

Displacement = Final Position - Initial Position Displacement = (+35 km) - (0 km) = +35 km

The '+' sign indicates the direction (in this case, to the right of the origin). Displacement has both magnitude (35 km) and direction (+). Therefore, it is a vector quantity.

Notice how different the values are! The car travelled a distance of 85 km, but its final displacement is only 35 km to the east (or in the positive direction).

{{VISUAL: diagram: The same road number line from 0 to 60 km. A dotted arrow traces the path of a car going from O to A (at 60 km) and then back to B (at 35 km). A separate, solid, bold arrow points directly from the starting point O to the final point B, labelled 'Displacement = 35 km'.}}

{{KEY: type=concept | title=Scalar and Vector Quantities | text=A scalar quantity is a physical quantity that has only magnitude (a numerical value) but no direction. Examples include distance, speed, mass, and time. A vector quantity is a physical quantity that has both magnitude and direction. Examples include displacement, velocity, force, and acceleration.}}

Worked Example: Motion on a Circular Path

A cyclist travels along a circular track of radius 70 m. They start at a point A and complete one full circle. What is their distance and displacement?

1. Calculate the Distance:

   • When the cyclist completes one full circle, the path they cover is equal to the circumference of the track.
   • Circumference C = 2πr
   • C = 2 × (22/7) × 70 m
   • C = 2 × 22 × 10 m = 440 m
   • So, the distance covered is 440 m.

2. Calculate the Displacement:

   • Displacement is the shortest distance between the initial and final positions.
   • The cyclist starts at point A and, after one full circle, ends up back at point A.
   • Since the initial and final positions are the same, the change in position is zero.
   • So, the displacement is 0 m.

This example perfectly illustrates a key idea: displacement can be zero even when the distance travelled is not.

Key Differences at a Glance

Let's summarize the crucial differences between distance and displacement in a table. This is a very common topic for exam questions.

{{KEY: type=exam | title=Common Question | text=A typical 3-mark question asks you to define distance and displacement and list three key differences between them. The table above provides a perfect structure for your answer.}}

Motion in a Straight Line — Part 2

Motion in a Straight Line — Part 2

In the previous section, we learned how to describe the position of an object and understood the difference between distance travelled and displacement. Now, we move to a deeper question: How do we describe how fast an object is moving? And more importantly, how do we describe the change in its speed?

These questions lead us to three fundamental concepts: average speed, average velocity, and average acceleration. Understanding these will help you not only solve numerical problems but also interpret real-world motion — from the sprint of an athlete to the journey of a car on a highway.

Average Speed

Imagine you are travelling from Delhi to Agra by car. The journey is 200 km long and takes you 4 hours. How would you describe how fast you travelled?

Average speed tells us the rate at which an object covers distance. It is defined as the total distance travelled divided by the total time taken.

{{KEY: type=definition | title=Average Speed | text=Average speed is the total distance travelled by an object divided by the total time taken to travel that distance. It is a scalar quantity and is always positive or zero.}}

Mathematically, we write:

Average speed = Total distance travelled / Total time taken

If we use symbols, let s be the total distance and t be the total time, then:

v_avg = s / t

The SI unit of average speed is metre per second (m/s). However, in everyday life, we often use kilometre per hour (km/h).

{{VISUAL: diagram: labeled diagram showing a car traveling along a curved road with distance markers and a clock showing elapsed time, illustrating the concept of average speed}}

Example 4.1

An athlete runs 160 m in 20 s. What is her average speed?

Solution:

• Total distance travelled, s = 160 m
• Total time taken, t = 20 s
• Average speed, v_avg = s / t = 160 m / 20 s = 8 m/s

So, the average speed of the athlete is 8 m/s.

{{KEY: type=exam | title=Common Trap | text=Average speed is calculated using TOTAL distance travelled, not displacement. Even if an object returns to its starting point, the average speed is not zero — but average velocity might be.}}

Average Velocity

Now, suppose the same athlete runs from point O to point B (40 m away) in 4 seconds, and then continues to point A (100 m from O) in another 6 seconds. What is her average velocity for the entire motion from O to A?

Average velocity is different from average speed because it takes into account direction. It is defined as the displacement divided by the time interval.

{{KEY: type=definition | title=Average Velocity | text=Average velocity is the displacement of an object divided by the time interval during which the displacement occurs. It is a vector quantity and can be positive, negative, or zero.}}

Mathematically:

Average velocity = Displacement / Time interval

Using symbols, if Δx represents displacement and Δt represents the time interval:

v = Δx / Δt

The SI unit of average velocity is also metre per second (m/s).

{{FORMULA: expr=v = Δx / Δt | symbols=v:average velocity (m/s), Δx:displacement (m), Δt:time interval (s)}}

Example 4.2

The athlete runs from O to A (100 m away) in 10 s. What is her average velocity?

Solution:

• Displacement, Δx = 100 m (in the positive direction)
• Time interval, Δt = 10 s
• Average velocity, v = Δx / Δt = 100 m / 10 s = 10 m/s

So, her average velocity is 10 m/s in the positive direction.

{{VISUAL: diagram: straight horizontal line with reference point O at left, points B and A marked to the right with distances labeled, showing displacement vector and time intervals for athlete's motion}}

Key Difference: Average Speed vs. Average Velocity

Let's compare the two with a real example. Suppose an athlete starts at O, runs to A (100 m away) in 10 s, then runs back to B (40 m from O) in another 6 s.

Notice that:

• Average speed is always the magnitude of how fast you cover ground, regardless of direction.
• Average velocity depends on your net change in position, so turning back reduces it significantly.

{{KEY: type=points | title=Speed vs. Velocity | text=- Average speed depends on total distance; average velocity depends on displacement.

• Average speed is always positive or zero; average velocity can be positive, negative, or zero.
• If an object returns to its starting point, average velocity is zero, but average speed is not.
• For motion in one direction without turning back, magnitude of average velocity equals average speed.}}

{{ZOOM: title=Why the Greek letter Δ (Delta)? | text=In physics and mathematics, Δ (capital Delta) is used to denote "change in" a quantity. So Δx means "change in position" (displacement), and Δt means "change in time" (time interval). This notation will become very useful as we study more complex motion in higher grades.}}

Average Acceleration

Now that you understand velocity, let's think about how velocity itself changes. When a car starts from rest and speeds up, or when a ball thrown upwards slows down, the velocity is changing. The rate at which velocity changes is called acceleration.

Average acceleration tells us how quickly the velocity of an object changes over a given time interval.

{{KEY: type=definition | title=Average Acceleration | text=Average acceleration is the change in velocity of an object divided by the time interval during which the change occurs. It is a vector quantity and can be positive, negative, or zero.}}

Mathematically:

Average acceleration = Change in velocity / Time interval

If we use symbols, let Δv be the change in velocity and Δt be the time interval:

a = Δv / Δt

where Δv = v − u (final velocity minus initial velocity).

{{FORMULA: expr=a = (v − u) / t | symbols=a:average acceleration (m/s²), v:final velocity (m/s), u:initial velocity (m/s), t:time interval (s)}}

The SI unit of acceleration is metre per second squared (m/s²).

{{VISUAL: chart: velocity-time graph showing a straight line with positive slope, labeled axes with velocity on y-axis and time on x-axis, illustrating constant positive acceleration}}

Example 4.3

A car's velocity increases from 10 m/s to 30 m/s in 5 seconds. What is its average acceleration?

Solution:

• Initial velocity, u = 10 m/s
• Final velocity, v = 30 m/s
• Change in velocity, Δv = v − u = 30 m/s − 10 m/s = 20 m/s
• Time interval, t = 5 s
• Average acceleration, a = Δv / t = 20 m/s / 5 s = 4 m/s²

So, the car's average acceleration is 4 m/s².

Positive and Negative Acceleration

• If velocity increases in the positive direction, acceleration is positive.
• If velocity decreases in the positive direction, acceleration is negative. Negative acceleration is often called retardation or deceleration.
• If velocity remains constant, acceleration is zero.

{{KEY: type=concept | title=Understanding Acceleration | text=Acceleration describes how fast velocity is changing. A positive acceleration means speeding up (in the reference direction), while a negative acceleration means slowing down. Even if an object is moving very fast, if its velocity is constant, its acceleration is zero.}}

Summary

In this section, you learned three powerful tools to describe motion quantitatively:

• Average speed: Total distance ÷ Total time (always positive, scalar).
• Average velocity: Displacement ÷ Time interval (can be positive, negative, or zero; vector).
• Average acceleration: Change in velocity ÷ Time interval (can be positive, negative, or zero; vector).

These concepts form the foundation of kinematics — the study of motion. In the next section, we will explore how to represent motion using graphs and equations, making it even easier to visualise and analyse how objects move.

Mastering these three concepts unlocks the language of motion — from everyday journeys to rocket launches, the same principles apply.

Plotting graph & Position-time graphs

Page 3: Plotting Graphs & Position-Time Graphs

Understanding motion mathematically is powerful, but graphs bring motion to life. A well-plotted graph can instantly tell you whether a car is speeding up, slowing down, or cruising steadily — all without a single calculation. In this section, you will learn how to plot motion graphs correctly and how to read and interpret position-time graphs to decode the story of an object's journey.

Plotting Graphs: A Step-by-Step Guide

Before we can interpret motion, we must learn to plot data accurately. A graph is a visual representation of the relationship between two quantities — in our case, position and time.

Preparing the Graph Paper

Graph paper is pre-divided into small squares, which helps you plot data with precision. The first step is to set up your axes:

1. Draw two perpendicular lines intersecting at a point called the origin (O).
2. The horizontal line is the x-axis, and the vertical line is the y-axis.
3. Decide which quantity goes on which axis. For motion, we typically plot time on the x-axis and position (or distance) on the y-axis.

{{VISUAL: diagram: labeled graph paper showing origin O, x-axis marked as Time, y-axis marked as Position, with grid squares visible}}

Choosing a Suitable Scale

The scale determines how much real-world quantity each division on the graph represents. A good scale should:

• Use the available space efficiently
• Make plotting and reading easy
• Allow all data points to fit comfortably

For example, if your time values range from 0 to 6 seconds and position from 0 to 120 metres, you might choose:

• x-axis scale: 5 divisions = 1 s
• y-axis scale: 5 divisions = 20 m

{{KEY: type=points | title=Tips for Choosing a Scale | text=- Avoid awkward numbers like 7 or 13 divisions per unit — stick to 1, 2, 5, or 10.

• Ensure the graph uses at least half the available page.
• Both axes need not have the same scale — choose independently for each quantity.}}

Plotting the Points

Once your axes and scale are ready, mark the data points:

1. Start at the origin (0 s, 0 m if applicable).
2. For each pair of values (e.g., time = 2 s, position = 40 m), find 2 s on the x-axis and 40 m on the y-axis.
3. Draw imaginary lines parallel to the axes from these points — their intersection is your data point.
4. Mark each point with a small dot or cross.

After plotting all points, connect them with a smooth line or curve, depending on the pattern.

{{KEY: type=concept | title=Graph ≠ Route Map | text=A position-time graph does NOT show the actual path taken by the object. It shows how the position coordinate changes over time with respect to a chosen origin. The object may have moved in a straight line, a curve, or zigzagged — the graph only tracks its position along one direction.}}

Position-Time Graphs: Reading the Story

A position-time graph is a powerful tool. Its shape, slope, and curvature reveal the nature of motion at a glance.

Straight Line = Constant Velocity

If the position-time graph is a straight line, the object is moving with constant velocity. In equal intervals of time, the object covers equal displacements.

Consider the data from our earlier example:

Plotting these points yields a perfectly straight line. Between any two 1-second intervals, the displacement is always 20 m — the velocity is constant.

{{VISUAL: chart: position-time graph showing a straight line passing through origin, with time on x-axis (0-6 s) and position on y-axis (0-120 m), labeled points at each second}}

{{KEY: type=definition | title=Constant Velocity | text=An object has constant velocity when its position-time graph is a straight line. The magnitude and direction of velocity remain unchanged throughout the motion.}}

Curved Line = Changing Velocity

If the graph is a curve, the velocity is not constant. The object is accelerating or decelerating.

Imagine a vehicle starting from rest and gradually speeding up:

Notice that in the first 2 seconds, it moves 1 m, but in the next 2 seconds it moves 3 m (4 − 1), and in the interval 8–10 s, it moves 9 m (25 − 16). The displacement in equal time intervals is increasing — the velocity is increasing, indicating acceleration.

{{VISUAL: chart: position-time graph showing a smooth upward curve (parabolic shape) starting from origin, with time on x-axis (0-12 s) and position on y-axis (0-40 m)}}

{{KEY: type=concept | title=Shape Reveals Motion Type | text=A straight line on a position-time graph indicates uniform motion (constant velocity), while a curve indicates non-uniform motion (changing velocity, i.e., acceleration). The steeper the curve becomes, the faster the velocity is increasing.}}

Extracting Information from the Graph

A position-time graph is not just a picture — it is a data mine. You can extract multiple physical quantities from it:

1. Position at Any Instant

Simply locate the time on the x-axis, trace upward to the curve, then move horizontally to read the position on the y-axis.

2. Displacement

The change in position between two instants gives displacement. If at t₁ = 2 s the position is 40 m and at t₂ = 5 s it is 100 m, the displacement is:

Displacement = Final position − Initial position = 100 m − 40 m = 60 m

3. Velocity from the Slope

The slope of the position-time graph gives the velocity. For a straight-line graph, the slope is constant, so velocity is constant. For a curved graph, the slope at any point gives the instantaneous velocity at that instant.

To calculate velocity from a straight-line graph:

{{FORMULA: expr=v = (x₂ − x₁) / (t₂ − t₁) | symbols=v:velocity (m/s), x₂:final position (m), x₁:initial position (m), t₂:final time (s), t₁:initial time (s)}}

Steeper slope → higher velocity. If the line is horizontal (slope = 0), the object is at rest.

{{KEY: type=exam | title=Common Exam Question | text=CBSE often asks: "Calculate the velocity from a given position-time graph." Always find the slope — pick two clear points on the line, calculate change in position divided by change in time. Show units in your answer.}}

{{ZOOM: title=Why does slope = velocity? | text=Velocity is defined as the rate of change of position, i.e., how fast position changes with time. Mathematically, rate of change is the slope of the graph. So the slope of the position-time graph is, by definition, velocity.}}

Remember: The position-time graph is your window into motion. A straight line means steady motion; a curve means the object is speeding up or slowing down. Master reading graphs, and you master motion.

Velocity-time graphs

Velocity-time Graphs

When you travel in a car, the speedometer needle doesn't stay fixed — it moves up and down as the car accelerates or slows down. A velocity-time graph is a powerful tool that captures this entire story of changing velocity on a single picture. Just as position-time graphs showed us where an object was at different moments, velocity-time graphs show us how fast an object was moving at each instant in time.

Understanding the Shape of Velocity-time Graphs

Let's explore three common scenarios you encounter in everyday motion:

Case 1: Constant Velocity

Imagine a car cruising on a straight highway at a steady speed of 72 km/h (or 20 m/s). Since the velocity isn't changing, the velocity-time graph is a horizontal straight line parallel to the time axis. The height of this line above the x-axis represents the magnitude of the constant velocity.

{{VISUAL: chart: velocity-time graph showing a horizontal line at 20 m/s parallel to the time axis, representing constant velocity motion}}

{{KEY: type=concept | title=Zero Acceleration from Horizontal Line | text=When the velocity-time graph is a horizontal straight line, the velocity is constant and the acceleration is zero. The slope of a horizontal line is zero, which directly tells us that there is no change in velocity with time.}}

Case 2: Uniformly Increasing Velocity

Now consider a car starting from rest (initial velocity = 0 m/s) and gradually picking up speed. Table 4.5 in your NCERT shows that in every 5-second interval, the velocity increases by 2.5 m/s. This gives us a straight line sloping upward from left to right. The uniform slope tells us that the velocity is increasing at a constant rate — in other words, the car is moving with constant acceleration.

Case 3: Uniformly Decreasing Velocity

What if a moving car applies brakes? Table 4.6 shows a car starting at 15.0 m/s and slowing down uniformly until it stops. The velocity-time graph is a straight line sloping downward. The car is still experiencing constant acceleration, but now the acceleration is in the opposite direction to the velocity (this is often called deceleration or retardation).

{{VISUAL: chart: three velocity-time graphs side by side showing (a) horizontal line for constant velocity, (b) upward sloping line for increasing velocity, (c) downward sloping line for decreasing velocity}}

The slope of a velocity-time graph reveals the nature of acceleration: zero slope means zero acceleration, positive slope means acceleration in the direction of motion, and negative slope means acceleration opposite to motion.

Calculating Acceleration from the Graph

Just as we found velocity from the slope of a position-time graph, we can find acceleration from the slope of a velocity-time graph. Let's see how.

The Slope Method

Consider the graph in Fig. 4.17d from your textbook. Pick any two points on the straight line — say point A at time t₁ with velocity u, and point B at time t₂ with velocity v.

The change in velocity is represented by the vertical distance BC = v − u

The change in time is represented by the horizontal distance CA = t₂ − t₁

Therefore, the slope of line AB is:

{{FORMULA: expr=a = (v − u) / (t₂ − t₁) | symbols=a:acceleration (m/s²), v:final velocity (m/s), u:initial velocity (m/s), t₁:initial time (s), t₂:final time (s)}}

{{KEY: type=definition | title=Acceleration from Velocity-time Graph | text=The slope of a straight line on a velocity-time graph gives the acceleration of the object. A positive slope indicates acceleration in the direction of velocity; a negative slope indicates acceleration opposite to the direction of velocity.}}

Worked Example

Let's calculate the acceleration for the car whose graph is shown in Fig. 4.17d between 10 s and 20 s:

• At point A (t = 10 s): velocity u = 5 m/s
• At point B (t = 20 s): velocity v = 10 m/s

Using our formula:

a = (10 m/s − 5 m/s) / (20 s − 10 s) = 5 m/s / 10 s = 0.5 m/s²

The positive value confirms that the car is speeding up.

For the decreasing velocity graph (Fig. 4.17e), following similar steps gives us a = −0.5 m/s². The negative sign is crucial — it tells us the acceleration is opposite to the direction of velocity, which is exactly what happens when brakes are applied.

{{KEY: type=exam | title=Sign Convention Matters | text=In CBSE exams, always mention the sign of acceleration and explain what it means. A negative acceleration doesn't always mean slowing down — it means acceleration is opposite to the chosen positive direction. Clear reasoning fetches full marks.}}

Calculating Displacement from the Graph

Here's a beautiful insight: the area enclosed between the velocity-time graph line and the time axis gives us the displacement of the object during that time interval. Let's understand why.

Case 1: Constant Velocity (Rectangular Area)

Look at Fig. 4.18a. For a car moving at constant velocity of 20 m/s for 6 seconds:

Area of rectangle OABC = base × height = OC × OA

Area = 6 s × 20 m/s = 120 m

Since velocity is constant, it equals the average velocity. Using the relation displacement = average velocity × time, we get:

Displacement = 120 m

{{VISUAL: chart: velocity-time graph with shaded rectangular area OABC between the graph line at 20 m/s and time axis from 0 to 6 seconds, showing how area represents displacement}}

{{KEY: type=concept | title=Area Under Graph Gives Displacement | text=The area enclosed between a velocity-time graph and the time axis for a given time interval equals the displacement of the object in that interval. For constant velocity, this area is a rectangle. For changing velocity, it may be a triangle or trapezium.}}

Case 2: Uniformly Changing Velocity (Triangular or Trapezoidal Area)

When velocity changes uniformly (as in Fig. 4.18b), the area between the graph and time axis forms a triangle (if starting from rest) or a trapezium (if starting with some initial velocity).

Suppose you want displacement between 10 s and 20 s for the accelerating car:

The shaded region is a trapezium with:

• Parallel sides (velocities): 5 m/s and 10 m/s
• Height (time interval): 10 s

Area of trapezium = ½ × (sum of parallel sides) × height

Displacement = ½ × (5 + 10) m/s × 10 s = ½ × 15 × 10 = 75 m

{{KEY: type=points | title=Steps to Find Displacement from Graph | text=- Identify the time interval for which displacement is needed.

• Identify the shape of the area enclosed (rectangle, triangle, trapezium).
• Apply the appropriate area formula with velocity as one dimension and time as the other.
• The numerical value of the area equals the magnitude of displacement in that interval.}}

{{ZOOM: title=Why Area Represents Displacement | text=Mathematically, displacement is the integral (sum) of velocity over time. Graphically, integration is represented by the area under a curve. Since velocity × time = distance (when velocity is constant), adding up tiny velocity × time strips under the graph gives total displacement — which is exactly what area calculation does.}}

Practical Insight

This area method works for any velocity-time graph, not just straight lines. For curved graphs (non-uniform acceleration), you can approximate the area by dividing it into small rectangles or triangles — a technique you'll explore in higher classes through calculus.

Understanding both slope and area of velocity-time graphs gives you complete control over analysing motion — you can extract acceleration, displacement, and even predict future motion from a single graph. This makes velocity-time graphs one of the most versatile tools in kinematics.

Kinematic Equations for Motion in a Straight

Page 5: Kinematic Equations for Motion in a Straight Line

4.3 Kinematic Equations for Motion with Constant Acceleration

When we watch a moving object — a car accelerating from a traffic light, a ball rolling down a slope — we often want to predict where it will be at a future moment or how fast it will be moving. To do this, we need a mathematical toolkit. In the special case where acceleration is constant, three elegant equations allow us to connect displacement, velocity, time, and acceleration.

Understanding Constant Acceleration

When acceleration is constant, the acceleration at every instant equals the average acceleration over any time interval. This simplification is powerful — it means we can use the definition of average acceleration directly.

Recall from Eq. (4.3c):

a = (v − u) / t

where u is the initial velocity at t = 0 s, v is the final velocity at time t, and a is the acceleration.

Rearranging this equation:

a t = v − u

v = u + a t

{{FORMULA: expr=v = u + a t | symbols=v:final velocity (m/s), u:initial velocity (m/s), a:acceleration (m/s²), t:time (s)}}

{{KEY: type=definition | title=First Equation of Motion | text=The equation v = u + a t relates final velocity, initial velocity, acceleration, and time for motion with constant acceleration. It allows us to find velocity at any instant if u and a are known.}}

Deriving the Second Equation: Displacement and Time

To find the displacement during time interval t, we turn to the velocity-time graph shown below.

{{VISUAL: diagram: velocity-time graph for motion with constant acceleration, showing initial velocity u at point A, final velocity v at point E, with rectangular area OACD and triangular area ABC shaded to represent displacement}}

The displacement s equals the area under the velocity-time curve. This area consists of:

• Rectangle OACD (representing motion at constant velocity u)
• Triangle ABC (representing the change in velocity due to acceleration)

Let us calculate:

1. Area of rectangle OACD = u × t
2. Area of triangle ABC = (1/2) × base × height = (1/2) × t × (v − u)

Total displacement:

s = u t + (1/2) t (v − u)

But from the first equation, v − u = a t. Substituting:

s = u t + (1/2) t (a t)

s = u t + (1/2) a t²

{{FORMULA: expr=s = u t + (1/2) a t² | symbols=s:displacement (m), u:initial velocity (m/s), a:acceleration (m/s²), t:time (s)}}

{{KEY: type=concept | title=Second Equation of Motion | text=The equation s = u t + (1/2) a t² gives displacement as a function of initial velocity, acceleration, and time. It shows that displacement depends quadratically on time when acceleration is present.}}

Deriving the Third Equation: Eliminating Time

Sometimes we know the velocities and acceleration, but not the time. Can we relate v, u, a, and s directly?

From the first equation:

t = (v − u) / a

Substitute this into the second equation:

s = u × [(v − u) / a] + (1/2) a × [(v − u) / a]²

Simplify step by step:

s = [u v − u²] / a + [a (v − u)²] / (2 a²)

s = [u v − u²] / a + [(v − u)²] / (2 a)

s = [2 u v − 2 u² + v² − 2 u v + u²] / (2 a)

s = [v² − u²] / (2 a)

Rearranging:

2 a s = v² − u²

v² = u² + 2 a s

{{FORMULA: expr=v² = u² + 2 a s | symbols=v:final velocity (m/s), u:initial velocity (m/s), a:acceleration (m/s²), s:displacement (m)}}

{{KEY: type=definition | title=Third Equation of Motion | text=The equation v² = u² + 2 a s connects final velocity, initial velocity, acceleration, and displacement without involving time. It is especially useful when time is unknown.}}

The Three Kinematic Equations: A Complete Toolkit

{{KEY: type=points | title=Conditions for Using Kinematic Equations | text=- Acceleration must be constant throughout the motion.

• Motion must be along a straight line.
• All quantities (u, v, a, s) must use consistent units.
• Signs indicate direction — negative a means deceleration or opposite direction.}}

{{ZOOM: title=Why Three Equations? | text=We have five physical quantities (u, v, a, s, t) and three independent equations. Each equation omits exactly one quantity. This is not a coincidence — it is a minimal, complete system for solving constant-acceleration problems.}}

Applying the Equations: Real-World Example

Example 4.8 (adapted): A car moving at 54 km/h applies brakes that cause an acceleration of −4 m/s². How far does it travel before stopping?

Step-by-Step Solution

Given:

• u = 54 km/h = 54 × (5/18) = 15 m/s
• a = −4 m/s² (negative because it opposes motion)
• v = 0 m/s (car stops)
• s = ?

Choose the right equation:

Since time t is unknown and not required, use the third equation:

v² = u² + 2 a s

Substitute:

0² = 15² + 2 × (−4) × s

0 = 225 − 8 s

8 s = 225

s = 28.125 m

Answer: The car travels approximately 28.1 m before stopping.

{{VISUAL: photo: car on a highway braking with skid marks visible on the road surface}}

{{KEY: type=exam | title=Common Exam Question Pattern | text=CBSE frequently asks 3-mark numerical problems requiring substitution in kinematic equations. Always write the given data, the equation used, and show substitution clearly. Watch for unit conversions — km/h to m/s is a common trap.}}

Connecting to Road Safety

Braking Distance and Safe Driving

The distance a vehicle travels after brakes are applied depends on:

• Initial velocity — higher speeds mean much longer stopping distances (quadratic relationship via s ∝ u²)
• Road surface — wet or icy roads reduce friction, lowering the magnitude of a
• Braking system — better brakes provide larger negative acceleration
• Driver reaction time — the time before brakes are even applied

{{VISUAL: diagram: comparison of braking distances for a car traveling at 30 km/h, 60 km/h, and 90 km/h, showing progressively longer stopping distances with labeled arrows}}

Modern vehicle-to-vehicle (V2V) communication systems can warn drivers of potential collisions by exchanging real-time velocity and position data.

{{KEY: type=concept | title=Quadratic Dependence on Speed | text=Because stopping distance s depends on v² (from v² = u² + 2 a s, with v = 0), doubling your speed quadruples your braking distance. This is why speed limits are strictly enforced in populated areas.}}

Summary and Looking Ahead

The three kinematic equations for constant acceleration are the foundation of motion analysis in a straight line. They allow us to predict where and how fast an object will be at any future moment — essential for everything from traffic engineering to spacecraft navigation.

In the next section, we move from straight-line motion to motion in a plane, where objects follow curved paths and new concepts like circular motion come into play.

Motion in a Plane & Summary & Quick Revision

Page 6: Motion in a Plane & Summary & Quick Revision

Motion in a Plane (Two-Dimensional Motion)

So far, we have explored motion along a straight line — one-dimensional motion. But what happens when an object moves in a plane? A vehicle overtaking another, a kicked football tracing a curve, or a satellite orbiting Earth — all these are examples of motion in two dimensions, also called motion in a plane.

In such motion, the object changes position in both horizontal and vertical directions simultaneously. The path traced is no longer a straight line but a curve or a circle.

{{VISUAL: diagram: top view of a car overtaking another car on a curved road, showing the curved path traced in a plane}}

Understanding Displacement in Circular Paths

Imagine a child sitting on a merry-go-round. As the merry-go-round spins, the child moves from point A to B to C along a curved path. Let us ask two important questions:

1. What is the distance travelled? It is the length of the curved arc ABC.
2. What is the displacement? It is the straight-line distance AC, from the starting point to the final position.

Clearly, distance and displacement are not equal in circular motion.

Now, if the child completes one full revolution and returns to the starting point, what is the displacement? It is zero, because the initial and final positions are the same. However, the distance travelled is the circumference of the circle, given by 2πR, where R is the radius.

{{KEY: type=concept | title=Distance vs Displacement in Circular Motion | text=For one complete revolution in a circular path, the distance travelled equals the circumference 2πR, but the displacement is zero because the object returns to its starting point.}}

Uniform Circular Motion

When an object moves along a circular path with constant (uniform) speed, the motion is called uniform circular motion.

Speed and Velocity in Uniform Circular Motion

Let us calculate the average speed of an object making one complete revolution. If the time taken for one revolution is T, then:

{{FORMULA: expr=v = 2πR / T | symbols=v:average speed (m/s), R:radius of circular path (m), T:time for one revolution (s)}}

But what about velocity? Since velocity is a vector (having both magnitude and direction), we must consider the direction of motion at every instant.

Key insight: In uniform circular motion, the speed is constant, but the direction of velocity changes continuously. At any point on the circle, the velocity is directed along the tangent to the circle at that point.

{{VISUAL: diagram: circular path with multiple arrows showing velocity vectors tangent to the circle at different points, illustrating how direction changes while magnitude remains constant}}

{{KEY: type=definition | title=Tangent to a Circle | text=A straight line that touches the circle at exactly one point is called a tangent to the circle at that point. The velocity of an object in circular motion is always along the tangent at its current position.}}

Is There Acceleration in Uniform Circular Motion?

You might think: "If the speed is constant, there is no acceleration." But remember, acceleration occurs whenever velocity changes — and velocity changes if either its magnitude or direction (or both) changes.

In uniform circular motion:

• The magnitude of velocity (speed) is constant.
• The direction of velocity changes continuously.

Therefore, uniform circular motion is accelerated motion, even though the speed remains constant. This is a subtle but powerful idea!

{{KEY: type=exam | title=Common Misconception | text=Students often assume constant speed means no acceleration. In exams, always remember: changing direction alone causes acceleration, even if speed is constant. This is tested in MCQs and conceptual questions.}}

A Thought Experiment: Releasing the Marble

Let us perform a simple mental experiment. Place a marble inside a ring and roll it so that it moves along the inner boundary. Now, suddenly lift the ring while the marble is moving. What happens?

Observation: The marble moves in a straight line along the tangent at the point where it was released.

Reason: Once released, the marble continues in the direction it was moving at that instant. There is no force keeping it in a circular path anymore. (You will explore the reason — involving forces and Newton's laws — in a later chapter.)

{{ZOOM: title=Motion in Space (Three Dimensions) | text=Real-world motion is often three-dimensional, such as a car climbing a mountain road, a bird flying, or an aircraft moving through air. You will study this in higher grades, building on the foundation of motion in one and two dimensions.}}

At a Glance: Chapter Summary

Let us quickly revise the key concepts from this chapter:

Position and Motion

• The position of an object is described by its distance and direction from a reference point at any instant of time.
• If the position changes with time, the object is said to be in motion with respect to that reference point.
• Rest and motion are relative: an object may be at rest with respect to one observer but in motion with respect to another.

{{KEY: type=points | title=Distance vs Displacement | text=- Distance is the total path length travelled (scalar quantity).

• Displacement is the shortest straight-line distance from initial to final position (vector quantity).
• Displacement can be zero even if distance is not zero (e.g., circular motion).}}

Speed and Velocity

• Average speed = Total distance / Total time
• Average velocity = Net displacement / Total time
• Uniform motion: speed and direction both remain constant.
• Non-uniform motion: speed or direction (or both) changes with time.

{{KEY: type=concept | title=Velocity is a Vector | text=Velocity has both magnitude and direction. Two objects can have the same speed but different velocities if they are moving in different directions. This distinction is critical in two-dimensional and circular motion.}}

Acceleration

Acceleration measures how quickly velocity changes with time.

{{FORMULA: expr=a = (v - u) / t | symbols=a:acceleration (m/s²), v:final velocity (m/s), u:initial velocity (m/s), t:time (s)}}

• Positive acceleration: velocity increases (e.g., a car speeding up).
• Negative acceleration (retardation or deceleration): velocity decreases (e.g., a car braking).
• Uniform acceleration: acceleration is constant over time.

{{KEY: type=points | title=Equations of Motion (Uniformly Accelerated Motion) | text=- v = u + at

• s = ut + ½at²
• v² = u² + 2as These three equations relate initial velocity, final velocity, acceleration, time, and distance travelled.}}

Graphical Representation

• Distance-time graph: Slope gives speed.
• Velocity-time graph: Slope gives acceleration; area under the curve gives displacement.

{{VISUAL: chart: two side-by-side graphs showing distance-time and velocity-time plots for uniform and non-uniform motion, with labels for slope and area}}

Motion in a Plane

• Motion in a plane (two-dimensional motion) includes circular motion, projectile motion, and more complex paths.
• Uniform circular motion: constant speed along a circular path, but velocity direction changes continuously.
• Even though speed is constant, uniform circular motion is accelerated because the direction of velocity changes.

{{KEY: type=exam | title=Exam Focus: Circular Motion | text=Examiners frequently ask why uniform circular motion is accelerated despite constant speed. Always explain: acceleration occurs due to continuous change in direction of velocity, not just change in magnitude.}}

Quick Revision Checklist

Before moving to the next chapter, ensure you can confidently:

✅ Define and distinguish between distance and displacement.
✅ Explain the difference between speed and velocity.
✅ Use the three equations of motion for uniformly accelerated motion.
✅ Interpret distance-time and velocity-time graphs.
✅ Calculate displacement, speed, velocity, and acceleration from graphs.
✅ Understand why uniform circular motion is accelerated motion.
✅ Apply the concept of relative motion to real-world examples.

Key Takeaway: Motion is the change in position with time. Understanding motion in one dimension lays the foundation for analysing more complex motion in two and three dimensions — and ultimately, for understanding the forces that cause this motion.

Congratulations! You have now mastered the fundamentals of describing motion. In the next chapter, you will discover why objects move the way they do — by exploring the concept of force and Newton's laws of motion.