You are viewing a free preview of this lesson.
Subscribe to unlock all 12 lessons in this course and every other course on LearningBro.
Kinematics is the branch of mechanics that describes how objects move, without worrying about why they move. It is the language of motion: if you can measure positions and times, you can reconstruct the entire story of a moving body — where it has been, how fast it was going, how its velocity changed, and what its future trajectory looks like.
This lesson is the foundation of OCR A-Level Physics A Module 3.1 (Motion). Every subsequent topic — dynamics, energy, momentum, simple harmonic motion, even gravitational fields — rests on a watertight understanding of the three quantities introduced here: displacement, velocity and acceleration.
Before defining anything, we must be clear about a distinction that OCR examiners love: the difference between scalar and vector quantities.
Exam Tip: OCR routinely asks "State the difference between distance and displacement" or "Explain why velocity is a vector". You must memorise the word direction. A common full-mark answer is: "Distance is a scalar (magnitude only); displacement is a vector (magnitude and direction from a reference point)."
The distinction is not pedantic. Consider an athlete running one full lap of a 400 m track. Their distance travelled is 400 m; their displacement (start to finish) is zero. Their average speed was non-zero; their average velocity was zero. Getting scalars and vectors the wrong way round is one of the most common exam errors in mechanics.
Displacement (symbol s or x) is the vector from a chosen origin to the object's current position. It is measured in metres (m).
A full displacement statement includes both size and direction:
The ball is 12 m north of the starting point. The lift is +8.5 m above the ground floor.
In one dimension, we use signs (+ and -) to indicate direction. In two dimensions we use either components (x, y) or a magnitude with a bearing.
A cyclist rides 3.0 km east, then 4.0 km north.
So the cyclist is 5.0 km from the start, in a specific direction, even though they travelled 7.0 km of road.
Velocity is the rate of change of displacement with time. It is a vector. The SI unit is metres per second (m s⁻¹).
Formally:
v = Δs / Δt (average velocity) v = ds/dt (instantaneous velocity)
An object travelling in a circle at a constant 10 m s⁻¹ has constant speed but changing velocity (the direction of motion keeps changing). This is the foundation of circular motion — later in the course you will see that a changing velocity means an acceleration, and hence a centripetal force.
A car drives 120 m east in 8.0 s, then 40 m west in 4.0 s.
Taking east as positive:
Now compute the average speed:
The two numbers differ because the car reversed direction. Always check which quantity the question asks for.
Acceleration is the rate of change of velocity with time. It is a vector. The SI unit is metres per second squared (m s⁻²).
a = Δv / Δt
A positive acceleration means velocity is increasing (in the chosen positive direction). A negative acceleration — often called deceleration — means the velocity is decreasing. Importantly, a negative acceleration does not necessarily mean the object is slowing down; it means the acceleration vector points in the negative direction. An object with negative velocity and negative acceleration is speeding up in the negative direction.
Common Exam Mistake: Confusing "negative acceleration" with "slowing down". Slowing down happens only when the velocity and acceleration vectors point in opposite directions. OCR mark schemes are strict on this.
A sprinter accelerates from rest to 9.0 m s⁻¹ in 1.8 s.
a = Δv / Δt = (9.0 − 0) / 1.8 = 5.0 m s⁻²
For comparison, the acceleration due to gravity at the Earth's surface is 9.81 m s⁻², so the sprinter's legs generate about half of g.
A displacement–time (s–t) graph is one of the two great tools of kinematics. On it:
flowchart LR
A[Displacement-Time Graph] --> B[Gradient]
B --> C[= Velocity]
A --> D[Shape]
D --> E[Straight line: constant v]
D --> F[Curve: accelerating]
D --> G[Horizontal: at rest]
If the graph is curved, draw a tangent at the point of interest and measure its gradient. Pick two widely separated points on the tangent to reduce reading error. The gradient of that tangent gives the instantaneous velocity at that instant.
The velocity–time (v–t) graph is even more powerful because it yields two physical quantities at once:
| Feature of v–t graph | Physical meaning |
|---|---|
| Horizontal line | Constant velocity, zero acceleration |
| Straight line with positive slope | Uniform acceleration |
| Curve | Non-uniform (changing) acceleration |
| Area above the time axis | Positive displacement |
| Area below the time axis | Negative displacement (motion reversed) |
A train accelerates uniformly from 0 to 30 m s⁻¹ in 20 s, then travels at 30 m s⁻¹ for a further 40 s, then decelerates uniformly to rest in 10 s. Find the total distance travelled.
The graph forms three regions:
Total displacement = 300 + 1200 + 150 = 1650 m
If the question had asked for average velocity, we would divide this by the total time: 1650 / 70 = 23.6 m s⁻¹.
Exam Tip: OCR questions on v–t graphs almost always ask you to calculate an area. Split the region into triangles, rectangles and trapezia and add them. The trapezium formula ½(a + b)h is especially useful for uniform acceleration.
A key distinction in A-Level mechanics:
The commonest real examples of uniform acceleration are:
Non-uniform acceleration occurs whenever resistive forces depend on velocity (drag, air resistance, viscous forces). A falling skydiver experiences non-uniform acceleration until reaching terminal velocity.
A ball is thrown vertically upwards at 15 m s⁻¹ from a balcony 20 m above the ground. Taking upwards as positive and ignoring air resistance:
On a v–t graph this appears as a straight line with gradient −9.81 m s⁻², starting at +15, crossing zero at t ≈ 1.53 s, and continuing negative until impact.
On an s–t graph the motion is a parabola opening downwards, peaking at maximum height.
This lesson covers OCR A-Level Physics A (H556), Module 3.1 (Motion):
In the next lesson we turn these definitions into a set of four algebraic relations — the SUVAT equations — that let us solve any problem involving uniform acceleration without drawing a graph.