What is motion?
A thing is in motion if its position changes with time. A thing is at rest if its position does not change.
To say where something is, we need a reference point (also called the origin). For example: "The school is 2 km north of the bus stand." Here the bus stand is the reference point.
Motion depends on who is watching. A passenger sitting in a moving bus is at rest for the person next to her, but in motion for a person standing on the road.
In this lesson we study motion in a straight line.
Distance and displacement
Distance is the total length of the path. It has only size (magnitude). A quantity with only size is a scalar.
Displacement is the shortest straight gap from the start point to the end point, with its direction. A quantity with size and direction is a vector.
- Distance is never negative. Displacement can be positive, negative or zero.
- If you come back to the start, displacement = 0, but distance is not 0.
- Displacement can never be bigger than distance.
Example: you walk 40 m east and then 15 m west. Distance = 55 m. Displacement = 25 m east.
SI unit of both: metre (m).
Uniform and non-uniform motion
In uniform motion an object covers equal distances in equal times, however small the time gaps. In the 3D, the dots dropped each second have equal gaps.
In non-uniform motion it covers unequal distances in equal times, like a bus in city traffic.
Speed and velocity
Speed = distance ÷ time. It tells how fast. It is a scalar.
Velocity = displacement ÷ time. It tells how fast and in which direction. It is a vector.
SI unit of both: m/s (also written m s⁻¹). Also used: km/h. To change km/h to m/s, multiply by 5/18. Example: 72 km/h = 72 × 5/18 = 20 m/s.
Average speed = total distance ÷ total time.
Average velocity = total displacement ÷ total time. If velocity changes at a steady rate, average velocity = (u + v) ÷ 2, where u is the start velocity and v the final velocity.
Velocity changes if the speed changes, or the direction changes, or both. A car going round a circular track at a steady speed still has a changing velocity, because its direction keeps changing (this is uniform circular motion).
Acceleration
Acceleration tells how fast the velocity changes.
a = (v − u) ÷ t
Here u = start (initial) velocity, v = final velocity, t = time taken. SI unit: m/s².
- If velocity increases, a is positive (in the direction of motion).
- If velocity decreases, a is negative. This is called retardation or deceleration, like a car braking.
- Uniform acceleration: velocity changes by equal amounts in equal times (a freely falling stone, a car speeding up steadily).
- Non-uniform acceleration: velocity changes by unequal amounts in equal times (a car in traffic).
Distance–time graphs
Put time on the x-axis and distance on the y-axis.
- Object at rest: a flat (horizontal) line.
- Uniform speed: a straight sloping line. The slope (rise ÷ run) = speed. A steeper line means a faster object.
- Speeding up: a curve that gets steeper and steeper (the grey line and the blue curve in the 3D).
To find speed from the graph, pick two points (t₁, s₁) and (t₂, s₂). Speed = (s₂ − s₁) ÷ (t₂ − t₁).
Velocity–time graphs
Put time on the x-axis and velocity on the y-axis.
- Constant velocity: a flat line. The area under it (a rectangle) = velocity × time = distance.
- Uniform acceleration: a straight sloping line. Its slope = acceleration.
- Uniform retardation: a straight line sloping down.
- Non-uniform acceleration: a curve.
The area under a v–t graph = displacement (distance for motion in one direction). For a car starting from rest and reaching 14 m/s in 7 s, area = ½ × 7 × 14 = 49 m.
Equations of motion (graphical method)
For motion in a straight line with uniform acceleration we get three equations. Take a v–t graph: the line starts at velocity u (at t = 0) and reaches v after time t.
1. Velocity–time relation: v = u + at
Slope of the line = a = (v − u) ÷ t. Rearranging gives v = u + at.
2. Position–time relation: s = ut + ½at²
Distance = area under the line = rectangle (u × t) + triangle (½ × t × (v − u)). Since v − u = at, s = ut + ½ × t × at = ut + ½at².
3. Position–velocity relation: v² = u² + 2as
The area is a trapezium: s = ½ × (u + v) × t. Put t = (v − u) ÷ a: s = (v + u)(v − u) ÷ 2a = (v² − u²) ÷ 2a. So v² = u² + 2as.
Tips: "starts from rest" means u = 0. "Comes to rest" or "stops" means v = 0. Braking means a is negative. Keep all units in m, s and m/s.
Try it: measure your own speed
Mark a 20 m straight path in your lane or ground (count about 26 long steps, or use a measuring tape). Ask a friend to time you with a phone stopwatch.
- Walk the 20 m. Note the time. Speed = 20 ÷ time.
- Now run it. Is your speed bigger?
- Ask your friend to call out your position every 2 seconds and draw your own distance–time graph. Is it a straight line?
In the 3D free-play step, predict first: with u = 0 and a = 2 m/s², how far does the car go in 5 s? Then check (answer: 25 m).
Key formulas and definitions
- Speed = distance ÷ time; velocity = displacement ÷ time
- Average speed = total distance ÷ total time
- Average velocity (uniform acceleration) = (u + v) ÷ 2
- Acceleration a = (v − u) ÷ t (unit m/s²)
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
- km/h → m/s: multiply by 5/18; m/s → km/h: multiply by 18/5
- Slope of s–t graph = speed; slope of v–t graph = acceleration; area under v–t graph = displacement
Worked examples
1. A girl walks 300 m north to a shop and then 100 m south to a friend's house. Find the distance and displacement.
Distance = 300 + 100 = 400 m. Displacement = 300 − 100 = 200 m towards north.
2. A bus covers 180 km in 3 hours. Find its average speed in km/h and in m/s.
Average speed = 180 ÷ 3 = 60 km/h. In m/s: 60 × 5/18 = 16.7 m/s (about).
3. A runner goes once round a circular track of radius 35 m in 44 s. Find the distance, displacement and average speed.
Distance = circumference = 2πr = 2 × 22/7 × 35 = 220 m. Displacement = 0 (he is back at the start). Average speed = 220 ÷ 44 = 5 m/s. Average velocity = 0 ÷ 44 = 0.
4. A scooter speeds up from 5 m/s to 15 m/s in 4 s. Find its acceleration.
a = (v − u) ÷ t = (15 − 5) ÷ 4 = 10 ÷ 4 = 2.5 m/s².
5. A car starts from rest and accelerates at 2 m/s² for 7 s. Find its final velocity and the distance covered.
u = 0, a = 2 m/s², t = 7 s. v = u + at = 0 + 2 × 7 = 14 m/s. s = ut + ½at² = 0 + ½ × 2 × 49 = 49 m. (Check with the v–t graph: triangle area = ½ × 7 × 14 = 49 m.)
6. A car moving at 72 km/h brakes and stops in 5 s. Find the retardation and the stopping distance.
u = 72 × 5/18 = 20 m/s, v = 0, t = 5 s. a = (0 − 20) ÷ 5 = −4 m/s² (retardation 4 m/s²). s = ut + ½at² = 20 × 5 + ½ × (−4) × 25 = 100 − 50 = 50 m.
7. A train at 10 m/s accelerates uniformly at 0.5 m/s² over 300 m. What is its velocity at the end?
Use v² = u² + 2as = 10² + 2 × 0.5 × 300 = 100 + 300 = 400. v = 20 m/s.
8. On a v–t graph, a trolley's velocity rises in a straight line from 0 to 6 m/s in 3 s, then stays at 6 m/s for 4 s. Find the acceleration in the first part and the total distance.
First part: slope = 6 ÷ 3 = 2 m/s². Distance = area = triangle + rectangle = ½ × 3 × 6 + 6 × 4 = 9 + 24 = 33 m.
Common mistakes
- Thinking distance and displacement are always equal. They are equal only when the object moves in one direction along a straight line.
- Forgetting to change km/h into m/s before using the equations. Multiply by 5/18.
- Using a positive a for braking. When velocity decreases, put a with a minus sign.
- Reading the slope of an s–t graph as acceleration. The slope of s–t is speed; the slope of v–t is acceleration.