Kinematics: Motion, Speed, Velocity, Acceleration, Graphs & Equations

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## **What is Kinematics?**
Kinematics is the **branch of physics that studies motion without considering the forces causing it**. It focuses on:
- **Position** (where an object is)
- **Displacement** (how far it moves from the starting point)
- **Speed** (how fast it moves)
- **Velocity** (speed + direction)
- **Acceleration** (how fast velocity changes)

### **Key Terms**
```figure
{
  "type": "number-line",
  "from": -10,
  "to": 10,
  "interval": {"from": -5, "to": 5, "openFrom": true, "openTo": false},
  "points": [
    {"x": 0, "label": "Start (O)"},
    {"x": 3, "label": "A (3 m)"},
    {"x": -2, "label": "B (-2 m)"}
  ],
  "caption": "A moves from O to A (displacement = +3 m), B moves from O to B (displacement = -2 m)."
}

• Distance: Total path length traveled (scalar, no direction). Example: If you walk 2 m east and 3 m north, your distance = 2 + 3 = 5 m.
• Displacement: Shortest distance from start to end (vector, has direction). Example: Same walk → displacement = √(2² + 3²) = 3.61 m northeast.
• Speed: Distance traveled per unit time (scalar). Formula: ( \text{Speed} = \frac{\text{Distance}}{\text{Time}} ) (units: m/s or km/h).
• Velocity: Displacement per unit time (vector). Formula: ( \text{Velocity} = \frac{\text{Displacement}}{\text{Time}} ) (units: m/s with direction).
• Acceleration: Rate of change of velocity (can be positive or negative). Formula: ( \text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Time}} ).

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Types of Motion

1. Uniform Motion (Constant Speed)

• Speed does not change over time.
• Graph: Straight line on a distance-time graph.
• Example: A car moving at 60 km/h on a highway.

2. Non-Uniform Motion (Variable Speed)

• Speed changes over time.
• Graph: Curved line on a distance-time graph.
• Example: A bus accelerating from rest.

3. Uniform Acceleration

• Acceleration remains constant.
• Example: A freely falling object (ignoring air resistance).
• Equations of Motion (for uniformly accelerated motion):
  1. ( v = u + at )
  2. ( s = ut + \frac{1}{2}at^2 )
  3. ( v^2 = u^2 + 2as )
  4. ( s = \frac{(u + v)}{2} \times t ) Where:
  • ( u ) = initial velocity
  • ( v ) = final velocity
  • ( a ) = acceleration
  • ( s ) = displacement
  • ( t ) = time

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Motion Graphs

Graphs help visualize motion. Three key types:

1. Distance-Time Graph

• Slope = Speed.
  • Steeper slope → higher speed.
  • Horizontal line → object at rest.
  • Curved line → acceleration.

2. Velocity-Time Graph

• Slope = Acceleration.
  • Horizontal line → constant velocity (no acceleration).
  • Area under the graph = displacement.

3. Acceleration-Time Graph

• Area under the graph = Change in velocity.

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Solved Examples

Example 1: Calculating Speed

A car travels 150 km in 2.5 hours. Find its average speed. Solution: [ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{150 \text{ km}}{2.5 \text{ h}} = 60 \text{ km/h} ]

Example 2: Using Equations of Motion

A bus starts from rest and accelerates at 2 m/s². Find its speed after 5 seconds and the distance covered. Solution:

1. Final velocity (( v )): [ v = u + at = 0 + (2 \times 5) = 10 \text{ m/s} ]
2. Distance covered (( s )): [ s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 5^2 = 25 \text{ m} ]

Example 3: Velocity-Time Graph Analysis

A velocity-time graph shows a straight line from (0,0) to (4,20). Find:

1. Acceleration.
2. Displacement in 4 seconds.

Solution:

1. Acceleration = slope of the graph = ( \frac{20 - 0}{4 - 0} = 5 \text{ m/s²} ).
2. Displacement = area under the graph = ( \frac{1}{2} \times 4 \times 20 = 40 \text{ m} ).

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Comparison Table: Speed vs. Velocity vs. Acceleration

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NEB Board-Style Questions

Short Answer (2 marks)

1. Define displacement. How is it different from distance? Answer: Displacement is the shortest distance from the starting point to the final position in a specified direction (vector). Distance is the total path length traveled (scalar). Example: Walking 3 m east and 4 m north → distance = 7 m, displacement = 5 m northeast.

2. A car moves with a constant speed of 25 m/s for 10 seconds. Calculate the distance covered. Answer: [ \text{Distance} = \text{Speed} \times \text{Time} = 25 \times 10 = 250 \text{ m} ]

Long Answer (5 marks)

3. A ball is thrown vertically upward with an initial velocity of 20 m/s. Calculate:

   • The time taken to reach maximum height.
   • The maximum height attained.
   • The velocity of the ball after 3 seconds.

   Solution:

   • At max height, ( v = 0 ). Using ( v = u + at ): [ 0 = 20 - 10t \implies t = 2 \text{ s} \quad (\text{since } a = -g = -10 \text{ m/s²}) ]
   • Max height (( s )): [ s = ut + \frac{1}{2}at^2 = 20 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 20 \text{ m} ]
   • Velocity after 3 s: [ v = 20 - 10 \times 3 = -10 \text{ m/s} \quad (\text{negative sign indicates downward direction}) ]

Graph-Based (3 marks)

4. Sketch a velocity-time graph for an object moving with:

   • Constant velocity of 5 m/s for 4 seconds.
   • Uniform acceleration of 2 m/s² for the next 3 seconds.

   Answer:

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Exam Tip

1. Units Matter: Always write units (m/s, m/s², km/h) in answers. NEB deducts marks for missing units!
2. Direction in Vectors: For velocity/acceleration, always state direction (e.g., "2 m/s north").
3. Graph Interpretation:
   • Distance-time: Slope = speed.
   • Velocity-time: Slope = acceleration; Area = displacement.
4. Equations of Motion: Memorize the 4 equations and know when to use each (e.g., if time is unknown, use $v^2=u^2+2as$).
5. Sign Conventions:
   • Upward motion: Positive velocity/acceleration.
   • Downward motion: Negative velocity/acceleration (e.g., free fall).
6. Practice Graphs: NEB often asks to sketch and interpret graphs. Practice drawing them neatly!

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