Distance-time and velocity-time graphs — how to find speed, acceleration and distance from any motion graph, with worked examples throughout.
Motion graphs appear on virtually every GCSE Physics paper. They're one of the most reliably tested skills — and one where students lose marks through predictable, avoidable errors. This guide goes through both types of motion graph in full detail, covering how to read them, how to calculate from them, and what each shape of graph actually means physically.
A distance-time graph shows how far an object has travelled from its starting point over time. Distance is on the y-axis, time is on the x-axis.
The gradient (slope) of a distance-time graph equals speed. This is the single most important fact about distance-time graphs.
❌ A distance-time graph cannot show negative values on the y-axis — distance cannot be negative (it's a scalar). If an object returns toward its start, the graph curves back but never goes below the x-axis. If the question involves displacement (a vector), a displacement-time graph CAN go negative. Check which quantity the graph is showing.
To find the speed at a specific instant on a curved graph, draw a tangent to the curve at that point and calculate the gradient of the tangent. This is the instantaneous speed. Examiners award marks for the tangent being correctly drawn and for the gradient calculation — show both clearly.
A velocity-time graph shows how an object's velocity changes over time. Velocity is on the y-axis, time on the x-axis. This is where most of the calculation marks are in motion graph questions.
The gradient of a velocity-time graph equals acceleration.
This is the second critical skill for velocity-time graphs. The area between the line and the x-axis gives the total distance (or displacement) travelled.
For a rectangle (constant velocity): area = length × height = time × velocity = distance. For a triangle (uniform acceleration from rest): area = ½ × base × height = ½ × time × final velocity. For a trapezium (uniform acceleration from a non-zero initial velocity): area = ½ × (v₁ + v₂) × t.
A car accelerates uniformly from 0 to 20 m/s in 5 seconds, then travels at 20 m/s for 10 seconds, then decelerates to rest in 4 seconds. Find the total distance travelled.
Section 1 (triangle): ½ × 5 × 20 = 50 m
Section 2 (rectangle): 10 × 20 = 200 m
Section 3 (triangle): ½ × 4 × 20 = 40 m
Total distance = 50 + 200 + 40 = 290 m
If the area under a velocity-time graph is an awkward shape, split it into rectangles and triangles. Count squares on the graph paper as a check — each square has an area of (x-scale unit) × (y-scale unit) in the appropriate units. Count all whole squares, then estimate the partial squares. This method is explicitly described in AQA mark schemes for "estimate the area" questions.
For uniform acceleration, the SUVAT equations relate the five quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). You only need two of them in GCSE Physics but they're very useful:
The AQA motion specification is at the AQA GCSE Physics specification page.
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