Speed, velocity and acceleration
Speed = distance ÷ time: , unit m/s — a scalar. Velocity is speed in a stated direction — a vector (same unit m/s); Cambridge requires the direction for full marks (10 m/s north ≠ 10 m/s south). Acceleration is the change in velocity per unit time: , unit m/s² — a vector; deceleration is negative acceleration. Mini-example: 4 m/s to 20 m/s in 8 s gives m/s². Cambridge free-fall value: m/s².
Reading d-t and s-t graphs
On a distance–time (d-t) graph: gradient = speed; a flat line = stationary; a curve = changing speed. On a speed–time (s-t) graph: gradient = acceleration; the area under the line = distance travelled; a flat line = constant speed. Read the y-axis label first to tell the two apart. Mini-example: an s-t triangle rising to 20 m/s over 4 s covers a distance = ½ × 4 × 20 = 40 m (the area), and its acceleration = 20 ÷ 4 = 5 m/s² (the gradient).
Drawn from real examiner reports.
s-t graph is not a d-t graph
On a speed-time graph the gradient is acceleration and the area is distance — not the other way round. Candidates calculate the gradient and call it distance, or read a horizontal line as "stationary". A flat s-t line means constant speed (zero acceleration); a flat d-t line means stationary. Read the y-axis label before you calculate.
June 2024 examiner report (Papers 31 and 33): both reports named treating the speed-time graph as a distance-time graph as the most common error in Q1. November 2024 mark scheme Paper 33 Q1(a)(i): "constant speed or moving with zero acceleration" is expected for a horizontal s-t section.
Acceleration: "rate" already means per time
Cambridge accepts "rate of change of velocity" OR "change in velocity per unit time" — but "rate of change of velocity per unit time" is marked wrong, because "rate" already means per unit time. Also reject "change in speed" (it must be velocity — direction matters). Defining deceleration only as "negative acceleration" earns compensatory credit, not full marks.
June 2024 examiner report (Paper 43 Q1a): explicitly names "rate of change of velocity per unit time" as an unacceptable definition of acceleration.
Forgetting the ½ on a triangle area
The area under a speed-time graph for uniform acceleration from rest is a triangle: distance = ½ × base × height, not base × height. Omitting the ½ doubles the answer. Sense-check with average speed: from rest to , the average is , so distance = . Example: ½ × 4 × 20 = 40 m, not 80 m.
June 2024 examiner report (Paper 31 Q1(a)(iii)): "Others calculated the answer to be 80 m by omitting the half for the area of a triangle." November 2024 mark scheme Paper 33 Q1(a)(ii) requires ½ × base × height for the accuracy mark.
Speed vs velocity — direction matters
Speed is a scalar (magnitude only); velocity is a vector (speed in a stated direction). Because velocity includes direction, an object moving in a circle at constant speed is still accelerating — its direction keeps changing. Defining velocity as "how fast something moves" or "a change in speed" scores no marks; you must state the direction element.
Say "constant speed", not "steady"
In "describe the motion" answers Cambridge wants precise words: "constant speed", "accelerating", "decelerating" or "stationary". "Steady speed", "it moves" or "it goes faster" are too vague and score zero. Give the type of motion, the numerical value where the graph is readable, and the direction where relevant.
Read the y-axis label first
Every motion Q1 is a graph. Read the y-axis: "distance / m" → d-t graph (gradient = speed, flat = stationary); "speed / m/s" → s-t graph (gradient = acceleration, area = distance). State what you are calculating ("area = distance") before substituting to earn the method mark.
Match the area shape to the motion
Choose the right area shape under a speed-time graph: triangle (from or to rest) = ½ × base × height; rectangle (constant speed) = base × height; trapezium (two non-zero speeds) = . The wrong shape is the usual reason a distance is out by a factor.
Describe motion: type, value, direction
For "describe the motion" marks, state three things: the type (constant speed / accelerating / decelerating / stationary), the numerical value where the graph is readable, and the direction where relevant. "The object moves" alone scores zero.
| Quantity | Symbol | Formula | Unit |
|---|---|---|---|
| Speed / velocity | |||
| Acceleration | |||
| Distance (from s-t graph) | = area under speed-time graph | ||
| Speed (from d-t graph) | = gradient of distance-time graph | ||
| Weight |
Where: = distance (m), = time (s), = initial speed (m/s), = final speed (m/s), = acceleration (), = mass (kg).
Cambridge standard: (gravitational field strength = ). Use this value unless the question states otherwise.
Scalars vs vectors:
Define speed.
A cyclist accelerates from to in .
(a) Calculate the acceleration of the cyclist. Give the unit of your answer.
(b) State whether this is an acceleration or a deceleration, and explain how you can tell from the sign of your answer.
