What this chapter is about
This chapter introduces you to the scientific way of describing how objects move. In everyday life, you see buses on roads, birds flying, and the Earth moving around the Sun. Physics gives you precise language and mathematical tools to describe all these movements.
You will learn the difference between distance and displacement, speed and velocity, and understand what acceleration means. The chapter builds your ability to use graphs to represent motion and to interpret what those graphs tell you. You will also learn the three equations of motion that connect displacement, velocity, acceleration and time for objects moving in a straight line with uniform acceleration.
After studying this chapter, you should be able to describe any motion using correct scientific terms, solve numerical problems involving moving objects, and read or draw distance-time and velocity-time graphs confidently.
Key ideas
- Rest and motion are relative: An object is at rest or in motion only with respect to a reference point. A passenger sitting in a moving train is at rest relative to another passenger but in motion relative to a tree outside.
- Distance is the total path length covered; displacement is the shortest straight-line distance from start to finish with direction. Distance is always positive; displacement can be zero, positive or negative.
- Speed is distance covered per unit time; velocity is displacement per unit time. Speed is a scalar (no direction); velocity is a vector (has direction).
- Uniform motion means equal distances are covered in equal time intervals. Non-uniform motion means the object covers unequal distances in equal time intervals.
- Acceleration is the rate of change of velocity. When velocity increases, acceleration is positive; when velocity decreases, acceleration is negative (also called retardation or deceleration).
- Equations of motion for uniformly accelerated motion along a straight line connect initial velocity (u), final velocity (v), acceleration (a), time (t) and displacement (s).
- Distance-time graphs: A straight line with slope indicates uniform speed; a curved line indicates changing speed. The slope of the graph gives speed.
- Velocity-time graphs: A horizontal line shows uniform velocity; a sloping line shows uniform acceleration. The area under the graph gives displacement.
Formulas and facts to remember
1. Speed = Distance / Time → tells how fast an object covers ground, ignoring direction.
2. Velocity = Displacement / Time → tells how fast and in which direction an object moves.
3. Acceleration = (Final velocity − Initial velocity) / Time = (v − u) / t → tells how quickly velocity changes.
4. First equation of motion: v = u + at → relates final velocity to initial velocity, acceleration and time.
5. Second equation of motion: s = ut + (1/2)at² → gives displacement when initial velocity, acceleration and time are known.
6. Third equation of motion: v² = u² + 2as → connects velocities, acceleration and displacement without needing time directly.
7. SI unit of speed and velocity: metre per second (m/s). SI unit of acceleration: metre per second squared (m/s²).
8. For an object moving in a circle and returning to the start, distance equals the circumference but displacement is zero.
Worked examples
### Example 1: Finding velocity and speed
A girl walks 400 m east and then 300 m north in 10 minutes. Find her average speed and average velocity.
Solution
Total distance = 400 m + 300 m = 700 m
Time = 10 min = 600 s
Average speed = 700 / 600 = 1.17 m/s (approximately)
Displacement = straight-line distance from start to finish. Using Pythagoras: Displacement = √(400² + 300²) = √(160000 + 90000) = √250000 = 500 m
Average velocity = 500 / 600 = 0.83 m/s (approximately), in the direction from start to finish (north-east direction).
### Example 2: Using equations of motion
A scooter starts from rest and accelerates uniformly at 2 m/s². Find its velocity and distance covered after 8 seconds.
Solution
Given: u = 0, a = 2 m/s², t = 8 s
Final velocity: v = u + at = 0 + 2 × 8 = 16 m/s
Displacement: s = ut + (1/2)at² = 0 × 8 + (1/2) × 2 × 64 = 0 + 64 = 64 m
The scooter reaches 16 m/s and covers 64 m.
### Example 3: Deceleration problem
A cyclist moving at 18 m/s applies brakes and comes to rest in 6 seconds. Find the acceleration and the distance covered while stopping.
Solution
Given: u = 18 m/s, v = 0, t = 6 s
Acceleration: a = (v − u) / t = (0 − 18) / 6 = −3 m/s² (negative sign shows retardation)
Distance: Using v² = u² + 2as 0 = 324 + 2 × (−3) × s 6s = 324 s = 54 m
The cyclist decelerates at 3 m/s² and travels 54 m before stopping.
Common mistakes
- Confusing distance with displacement → Remember: distance is path length (always positive); displacement is shortest straight-line separation with direction.
- Using speed and velocity interchangeably → Speed has no direction; velocity must include direction.
- Forgetting to convert minutes to seconds in problems → Always use SI units (seconds, metres) before substituting in formulas.
- Taking acceleration as always positive → Acceleration is negative when velocity decreases; include the sign in calculations.
- Misreading graph slopes → On a distance-time graph, slope gives speed; on a velocity-time graph, slope gives acceleration; area under velocity-time graph gives displacement.
Quick revision
- Rest and motion depend on the reference point chosen.
- Distance ≥ |Displacement|; they are equal only for straight-line motion without turning back.
- v = u + at; s = ut + (1/2)at²; v² = u² + 2as — memorise and practise these three equations.
- Slope of distance-time graph = speed; slope of velocity-time graph = acceleration.
- Area under velocity-time graph = displacement covered.
- Always use SI units and watch the sign of acceleration.