What Is Motion? A Visual Guide for OxfordAQA IGCSE Physics Students

Motion, within OxfordAQA IGCSE Physics, is the description of how an object's position changes over time, expressed through the linked quantities of distance, displacement, speed, velocity and acceleration, and represented graphically through distance-time and velocity-time graphs. It sits at the foundation of the whole Forces and their effects section, since resultant force, Newton's laws and momentum all build directly on the definitions established here.

This guide sets out what is motion igcse physics expects with precision: a clear oxfordaqa igcse physics definition of every quantity first, then the graphical and numerical techniques that appear consistently across past papers. Treat it as your reference for oxfordaqa igcse motion, with motion explained from first principles rather than assumed.

Key Facts

Speed and velocity are related but distinct: velocity has a direction, speed does not.
A distance-time graph's gradient gives speed; a velocity-time graph's gradient gives acceleration.
The area under a velocity-time graph gives the distance travelled.
Acceleration is the rate of change of velocity, and it can occur through a change in direction, even at constant speed.

Distance, Displacement, Speed and Velocity

If an object moves in a straight line, its distance from a fixed reference point can be represented on a distance-time graph, with time on the horizontal axis and distance on the vertical axis. The speed of the object at any point can be calculated from the gradient of this graph: a steeper gradient means a greater speed, a flat, horizontal section means the object is stationary.

Velocity is a more precise quantity than speed, because it specifies direction as well as magnitude. The velocity, v, of an object is defined by the equation v = s ÷ t, where s is the displacement, the distance travelled in a specified direction, and t is the time taken. This same equation is also used to calculate average speed for objects whose motion is not uniform, that is, objects that speed up or slow down during their journey, in which case the result gives the average rather than the instantaneous value.

It's worth being clear about the distinction between distance and displacement, since the specification uses displacement specifically in the velocity equation. Distance is the total length of the path travelled, regardless of direction, while displacement is the straight-line distance from the starting point to the finishing point, in a specified direction. A runner who completes one full lap of a 400 m track has travelled a distance of 400 m, but their displacement is zero, because they have returned to their starting point. This is precisely why velocity, calculated from displacement, can be different from speed, calculated from distance, even for the same journey.

Worked example. A cyclist travels 900 m in 60 s. Calculate her average speed.
v = s ÷ t = 900 ÷ 60 = 15 m/s.

Resultant Force and Acceleration

Motion cannot be fully understood in isolation from force, since it is a non-zero resultant force that causes an object to accelerate. Acceleration is defined as the rate of change of velocity, given by the equation a = Δv ÷ t, where Δv is the change in velocity and t is the time taken for that change to occur. Deceleration is simply a negative acceleration, describing an object slowing down.

A subtlety worth stating precisely: an object can accelerate purely by changing direction, even while travelling at a constant speed. A car travelling around a bend at a steady 30 km/h is accelerating, because its velocity, which includes direction, is continuously changing, even though its speed is not. This distinction is one of the most frequently tested ideas in this part of the specification, precisely because it contradicts the everyday, informal use of the word "acceleration" to mean only "speeding up."

It helps to separate the two situations in which an object accelerates: a change in speed, and a change in direction. Both count as acceleration under the physics definition, but they feel quite different intuitively, and questions often test whether you can identify acceleration in a scenario that involves only a change in direction, such as a satellite in a circular orbit or a car navigating a roundabout, where the speed shown on the speedometer might not change at all.

Worked example. A car accelerates from 10 m/s to 28 m/s in 6 s. Calculate its acceleration.
Δv = 28 - 10 = 18 m/s.
a = Δv ÷ t = 18 ÷ 6 = 3 m/s².

Newton's Laws, in Brief

Motion and force are formally connected through Newton's three laws, and while the full detail belongs to the resultant forces topic, the essential statements are worth having alongside the equations above, since exam questions on motion frequently draw on them together.

Law	Statement
First	If the resultant force on an object is zero, a moving object continues at the same velocity, and a stationary object remains at rest
Second	F = m × a; a non-zero resultant force causes acceleration in the direction of that force
Third	When two objects interact, the forces they exert on each other are equal in magnitude and opposite in direction

Reading Motion Graphs

Both distance-time and velocity-time graphs appear regularly, and it's essential to know exactly what each axis, gradient and area represents, since the two graph types encode different information from superficially similar-looking lines.

Graph	Gradient represents	Area under graph represents	Flat, horizontal section means
Distance-time	Speed	Not physically meaningful	Object is stationary
Velocity-time	Acceleration	Distance travelled	Object moving at constant velocity, zero acceleration

Worked example. A velocity-time graph shows an object accelerating uniformly from rest to 20 m/s over 5 s, then travelling at a constant 20 m/s for a further 10 s. Calculate the total distance travelled.
Distance during acceleration = area of triangle = ½ × base × height = ½ × 5 × 20 = 50 m.
Distance at constant velocity = area of rectangle = base × height = 10 × 20 = 200 m.
Total distance = 50 + 200 = 250 m.

Breaking a velocity-time graph into simple triangular and rectangular sections, and finding the area of each separately before summing them, is the standard, reliable technique for this kind of question, and it's worth practising until it becomes automatic.

The same triangle-and-rectangle approach extends naturally to more complex graphs made up of several straight-line sections, for example an object that accelerates, then travels at constant velocity, then decelerates to a stop. Treat each section of the graph separately, find the area under each one using the appropriate shape, whether triangle, rectangle or trapezium, and sum the results to find the total distance travelled across the whole journey.

An International Comparison

Students who have previously studied physics elsewhere in Europe will find the definitions of speed, velocity and acceleration entirely familiar, since these are internationally standard quantities defined identically wherever they're taught. What sometimes needs adjustment is exam technique specific to this specification: OxfordAQA questions on motion place particular emphasis on the graphical interpretation of distance-time and velocity-time graphs, and on precise written explanations of why acceleration occurs during a change of direction at constant speed. If your previous study leaned more heavily on formula-based calculation than graph interpretation, it's worth deliberately practising the graphical questions on past papers, since the underlying physics is familiar but the assessment style may not be.

A Worked Example Combining Distance and Velocity-Time Graphs

Worked example. A cyclist accelerates uniformly from rest to 8 m/s over 4 s, maintains 8 m/s for 12 s, then decelerates uniformly to rest over 4 s. Calculate the total distance travelled, and the average speed for the whole journey.
Distance during acceleration = ½ × 4 × 8 = 16 m.
Distance at constant velocity = 12 × 8 = 96 m.
Distance during deceleration = ½ × 4 × 8 = 16 m.
Total distance = 16 + 96 + 16 = 128 m.
Total time = 4 + 12 + 4 = 20 s.
Average speed = total distance ÷ total time = 128 ÷ 20 = 6.4 m/s.

Notice that the average speed, 6.4 m/s, is lower than the constant cruising speed of 8 m/s, because the acceleration and deceleration phases pull the average down. This is a useful sanity check to apply to any similar question: your calculated average should always sit somewhere between the lowest and highest speeds reached during the journey, and if it doesn't, that's a strong signal an arithmetic error has crept in somewhere.

Common Mistakes

Treating speed and velocity as interchangeable terms, and consequently missing marks on questions that specifically ask for a change in direction to be identified as a form of acceleration.
Reading a flat section of a velocity-time graph as "not moving," when it actually represents motion at a constant, non-zero velocity.
Confusing what a graph's gradient represents with what the area beneath it represents, particularly under time pressure.
Forgetting to break a non-uniform velocity-time graph into separate triangular and rectangular sections before calculating distance, and instead attempting a single incorrect calculation across the whole graph.

Self-Check Questions

Explain the difference between speed and velocity.
A runner covers 400 m in 50 s at a constant speed. Calculate her speed.
Explain why an object moving at constant speed around a circular track is still accelerating.
A cyclist accelerates from 4 m/s to 12 m/s in 4 s. Calculate the acceleration.
Sketch a velocity-time graph for an object that accelerates uniformly from rest to 15 m/s over 3 s, and use it to calculate the distance travelled in that time.
State what a flat, horizontal section of a distance-time graph indicates about an object's motion.

This is oxfordaqa igcse physics explained the way motion actually gets tested: through these definitions, worked examples and graph techniques together, forming the foundation the rest of Forces and their effects builds on. Keep these oxfordaqa igcse physics notes alongside your own notes from class, so time spent making these ideas genuinely secure pays off well beyond this topic alone.