Motion - Distance, displacement, speed, velocity, acceleration, and equations of motion

Hello dear student! Welcome to this comprehensive guide on Motion.

In physics, everything around us—from the smallest atom to the giant planets—is in constant movement. Whether you are walking to school, riding a bicycle, or watching a bird fly, you are observing motion in action.

In this tutorial, we will break down the fundamental concepts of motion into simple, bitesize ideas using everyday examples, easy-to-remember formulas, and step-by-step problem-solving. Grab your notebook, and let's dive in!

---

1. What is Motion & The Reference Point

Before we define speed or acceleration, let's answer a simple question: How do you know something is moving?

You know an object is moving when its position changes with time. But relative to what?

The Reference Point (Origin)

Imagine you are sitting inside a moving train.

To a passenger sitting next to you, you are at rest.

To a person standing on the railway platform, you are in motion.

Thus, to describe the position of an object, we need to specify a fixed point called the Reference Point (also called the Origin).

Definition: An object is said to be in motion if it changes its position with respect to a fixed reference point over time.

---

2. Distance vs. Displacement: The Tale of Two Paths

Suppose you leave your house ($A$), walk $3\text{ km}$ to buy an ice cream at ($B$), and then walk $4\text{ km}$ to your friend's home at ($C$).

```

A (Home) ---------------> B (Ice Cream Shop)

| 4 km

C (Friend's House)

```

How far did you travel?

Distance

What is it? The total length of the actual path covered by a moving object, regardless of direction.

In our example: Path $AB + BC = 3\text{ km} + 4\text{ km} = 7\text{ km}$.

Key Characteristics:

It has magnitude (value) only, no direction (Scalar quantity).

It can never be zero or negative for a moving object.

SI Unit: Meter ($\text{m}$).

Displacement

What is it? The shortest straight-line distance from the initial position to the final position, measured along with direction.

In our example: The direct straight line from $A$ to $C$ (using Pythagoras theorem: $\sqrt{3^2 + 4^2} = 5\text{ km}$).

Key Characteristics:

It has both magnitude and direction (Vector quantity).

It can be zero, positive, or negative!

SI Unit: Meter ($\text{m}$).

Teacher's Tip — Why Displacement can be Zero:

If you go to school in the morning and return home in the evening along the same route:

* Distance covered = $\text{Home to School} + \text{School to Home} = 2 \times \text{Distance}$

* Displacement = $0\text{ m}$ (because your starting point and ending point are the same!).

---

Comparison Table: Distance vs. Displacement

Feature	Distance	Displacement
Definition	Actual total path length traveled.	Shortest distance between initial and final points.
Quantity Type	Scalar (Magnitude only)	Vector (Magnitude + Direction)
Can it be zero?	No (for a moving body)	Yes (if initial & final points are same)
Value	Always $\ge$ Displacement	Always $\le$ Distance

---

3. Speed and Velocity: How Fast Are You Moving?

A. Speed (How fast?)

Speed tells us how fast an object is moving.

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$

SI Unit: Meters per second ($\text{m/s}$ or $\text{m}\cdot\text{s}^{-1}$).

Type: Scalar quantity.

Average Speed

In real life, we rarely travel at a constant speed. We slow down at traffic signals and speed up on clear roads. Hence, we calculate Average Speed:

$$\text{Average Speed} = \frac{\text{Total Distance Traveled}}{\text{Total Time Taken}}$$

---

B. Velocity (How fast and in which direction?)

Velocity is simply speed given in a specific direction. It is the rate of change of displacement.

$$\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}$$

SI Unit: Meters per second ($\text{m/s}$).

Type: Vector quantity.

Velocity changes if either the speed changes or the direction of motion changes.

Average Velocity

When velocity changes at a uniform rate, we can calculate Average Velocity as:

$$\text{Average Velocity} = \frac{u + v}{2}$$

Where:

$u = \text{Initial velocity}$

$v = \text{Final velocity}$

---

C. Uniform vs. Non-Uniform Motion

Uniform Motion: An object covers equal distances in equal intervals of time (e.g., a car cruising at a constant $60\text{ km/h}$ on a straight highway).

Non-Uniform Motion: An object covers unequal distances in equal intervals of time (e.g., a bus moving through city traffic).

---

4. Acceleration: Changing Gears!

Have you ever felt pushed back into your seat when a car speeds up suddenly? That feeling is caused by acceleration.

Definition: Acceleration is defined as the rate of change of velocity with time.

$$\text{Acceleration } (a) = \frac{\text{Change in Velocity}}{\text{Time Taken}} = \frac{\text{Final Velocity } (v) - \text{Initial Velocity } (u)}{t}$$

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

SI Unit: Meter per second squared ($\text{m/s}^2$ or $\text{m}\cdot\text{s}^{-2}$).

Types of Acceleration:

Positive Acceleration: When velocity increases over time (in the direction of motion).

Negative Acceleration (Deceleration / Retardation): When velocity decreases over time (e.g., applying brakes).

Zero Acceleration: When an object moves with constant velocity ($v = u$).

---

5. The Three Equations of Motion

When an object moves along a straight line with uniform acceleration, its motion can be described using three simple algebraic equations. These are the core tools for solving numerical problems in NCERT Class 9 physics!

The Notation Rules:

$u$ = Initial velocity ($\text{m/s}$)

$v$ = Final velocity ($\text{m/s}$)

$a$ = Acceleration ($\text{m/s}^2$)

$t$ = Time taken ($\text{s}$)

$s$ = Distance / Displacement ($\text{m}$)

---

First Equation of Motion (Velocity-Time Relation)

$$v = u + at$$

Use this when: You need to find final velocity, initial velocity, acceleration, or time, and distance is not involved.

---

Second Equation of Motion (Position-Time Relation)

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

Use this when: You need to calculate the distance traveled given initial velocity, time, and acceleration.

---

Third Equation of Motion (Position-Velocity Relation)

$$v^2 - u^2 = 2as \quad \text{or} \quad v^2 = u^2 + 2as$$

Use this when: Time ($t$) is not given in the problem!

---

6. Quick Formula Cheat-Sheet

```

Distance = Speed × Time

Speed = Distance / Time

Velocity = Displacement / Time

Acceleration (a) = (v - u) / t

1st Equation of Motion: v = u + at

2nd Equation of Motion: s = ut + ½ at²

3rd Equation of Motion: v² - u² = 2as

Unit Conversion: km/h to m/s -> Multiply by (5 / 18)

```

---

7. Practice Questions with Detailed Step-by-Step Solutions

Let's test our understanding with 3 classic textbook numerical problems. Try solving them on your own before reading the solution!

---

Question 1 (Distance & Displacement)

An athlete completes one round of a circular track of diameter $200\text{ m}$ in $40\text{ seconds}$. What will be the distance covered and the displacement at the end of $2\text{ minutes } 20\text{ seconds}$?

Solution:

Step 1: Write down given values.

Diameter of circular track, $D = 200\text{ m}$

Radius of track, $r = \frac{D}{2} = 100\text{ m}$

Time for 1 round = $40\text{ s}$

Total time = $2\text{ min } 20\text{ s} = (2 \times 60) + 20 = 140\text{ s}$

Step 2: Find the number of rounds completed in $140\text{ seconds}$.

$$\text{Number of rounds} = \frac{\text{Total Time}}{\text{Time for 1 round}} = \frac{140}{40} = 3.5 \text{ rounds}$$

Step 3: Calculate Total Distance.

Circumference of track = $2\pi r = 2 \times \frac{22}{7} \times 100 = \frac{4400}{7}\text{ m}$

$$\text{Total Distance} = \text{Rounds} \times \text{Circumference}$$

$$\text{Total Distance} = 3.5 \times \frac{4400}{7} = \frac{7}{2} \times \frac{4400}{7} = 2200\text{ m}$$

Step 4: Calculate Displacement.

After $3$ full rounds, the athlete is back at the starting point (Displacement = $0$).

The remaining $0.5$ (half) round puts the athlete exactly at the diametrically opposite point.

Therefore, the shortest distance from start to end position is the diameter of the circle.

$$\text{Displacement} = \text{Diameter} = 200\text{ m}$$

Final Answer:

Distance covered: $2200\text{ m}$

Displacement: $200\text{ m}$

---

Question 2 (Acceleration & Unit Conversion)

A bus decreases its speed from $80\text{ km/h}$ to $60\text{ km/h}$ in $5\text{ seconds}$. Find the acceleration of the bus.

Solution:

Step 1: Write down given values and convert units to SI ($\text{m/s}$).

Initial velocity, $u = 80\text{ km/h} = 80 \times \frac{5}{18} = \frac{200}{9}\text{ m/s} \approx 22.22\text{ m/s}$

Final velocity, $v = 60\text{ km/h} = 60 \times \frac{5}{18} = \frac{150}{9}\text{ m/s} \approx 16.67\text{ m/s}$

Time, $t = 5\text{ s}$

Step 2: Apply the acceleration formula.

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

$$a = \frac{\frac{150}{9} - \frac{200}{9}}{5} = \frac{-\frac{50}{9}}{5} = -\frac{50}{9 \times 5} = -\frac{10}{9}\text{ m/s}^2$$

$$a = -1.11\text{ m/s}^2$$

Final Answer:

The acceleration of the bus is $-1.11\text{ m/s}^2$ (The negative sign indicates deceleration/slowing down).

---

Question 3 (Equations of Motion)

A train starting from rest attains a velocity of $72\text{ km/h}$ in $5\text{ minutes}$. Assuming that the acceleration is uniform, find:

The acceleration of the train.

The distance traveled by the train for attaining this velocity.

Solution:

Step 1: Extract given information and convert to SI units.

Starting from rest $\implies$ Initial velocity, $u = 0\text{ m/s}$

Final velocity, $v = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$

Time, $t = 5\text{ minutes} = 5 \times 60 = 300\text{ s}$

---

Part 1: Find Acceleration ($a$)

Using the first equation of motion:

$$v = u + at$$

$$20 = 0 + a \times 300$$

$$a = \frac{20}{300} = \frac{1}{15}\text{ m/s}^2 \approx 0.067\text{ m/s}^2$$

---

Part 2: Find Distance Traveled ($s$)

Using the third equation of motion:

$$v^2 - u^2 = 2as$$

$$(20)^2 - (0)^2 = 2 \times \left(\frac{1}{15}\right) \times s$$

$$400 = \frac{2}{15} \times s$$

$$s = \frac{400 \times 15}{2} = 200 \times 15 = 3000\text{ m} = 3\text{ km}$$

Final Answer:

Acceleration: $\frac{1}{15}\text{ m/s}^2$ (or $0.067\text{ m/s}^2$)

Distance traveled: $3000\text{ m}$ (or $3\text{ km}$)

---

Final Words of Encouragement

You've done a fantastic job going through the fundamentals of motion! Remember:

Always convert units to SI units ($\text{m}$, $\text{s}$, $\text{m/s}$, $\text{m/s}^2$) before putting values into equations.

Read numericals carefully to spot clues like "starts from rest" ($u = 0$) or "comes to rest/brakes applied" ($v = 0$).

Keep practicing, stay curious, and happy learning!