Distance–time graphs: gradient = speed
On a distance–time (d-t) graph the gradient = speed; a steeper line = higher speed. A flat section = stationary (distance constant); a straight slope = constant speed; a curve = changing speed (steepening → speeding up, flattening → slowing). Find speed from two well-separated points: (rise/run). E.g. from (0 s, 0 m) to (10 s, 60 m): m/s. Units: m, s, m/s.
Velocity–time graphs: gradient and area
On a velocity–time (v-t) graph the gradient = acceleration, : positive = speeding up, negative = slowing, flat = constant velocity (zero acceleration). The area under the line = distance travelled — rectangle , triangle base height, trapezoid ; for a curve, estimate by counting grid squares or dividing it into strips. Units: m/s, s, m/s², m.
Match the motion to the graph shape
Each motion has a signature shape. Stationary: d-t horizontal; v-t flat at . Constant speed: d-t straight sloped; v-t flat above zero. Uniform acceleration: d-t upward curve (steepening); v-t straight positive gradient. Uniform deceleration: d-t curve flattening; v-t straight negative gradient. Free fall: v-t straight line of gradient m/s². Critical: a flat v-t line means constant velocity, NOT stopped — stopped is a flat line at .
Drawn from real examiner reports.
Tangent read as a point, not a gradient
To find acceleration on a curved v-t graph, draw a tangent at the point, take TWO well-separated points ON the tangent and compute = rise/run. The frequent error is to read and at that single point and divide — that is velocity ÷ time since the start, not the tangent's gradient, so it scores zero.
June 2024 Paper 1P Q12(a): most candidates attempted to draw a tangent appropriately but few then used it correctly to determine the acceleration. Many candidates simply read the velocity and time from that point on the curve and substituted them into the acceleration equation, which was not credited.
SUVAT used on a curved v-t graph
SUVAT (, , ) is valid ONLY for constant acceleration — a straight v-t line. On a curved v-t graph the acceleration changes, so SUVAT does not apply: find distance from the area under the curve (strips or grid squares). Writing for a curve earns no credit.
June 2024 Paper 1P Q12(b): few candidates appreciated that they needed to determine the area under the curve to find the distance travelled. Many candidates used $v^2 = u^2 + 2as$ and received no credit, as this formula is invalid for non-constant acceleration.
Constant velocity stated without forces
When a v-t graph goes flat, "constant velocity" earns just 1 mark. For full marks explain the force balance: the resultant force is zero / driving force = resistive force / forces are balanced, so by the acceleration is zero. Stopping at "it moves at a constant speed" caps you at one mark.
June 2024 Paper 1P Q12(c): most candidates recognised that the car was moving at constant velocity after 80 seconds, but only half of all candidates gained further credit for attempting to explain this. Only the most able candidates gained full marks by explaining that forces were balanced.
Flat d-t line vs flat v-t line
A flat line means different things on the two graphs. On a d-t graph a horizontal line = stationary (not moving). On a v-t graph a horizontal line = constant velocity (still moving, zero acceleration). Read the y-axis label first: distance → d-t; velocity → v-t. Confusing these is the commonest graph-interpretation error.
Gradient unit left as m/s not m/s²
The gradient unit is the y-axis unit ÷ the x-axis unit. For a v-t graph that is m/s ÷ s = m/s², not m/s. Quoting an acceleration in m/s (the speed unit) loses the unit mark even when the number is right. For a d-t graph the gradient unit is m ÷ s = m/s.
Distance vs displacement under a v-t graph
The area under a v-t graph is the distance travelled (m). Call it displacement only if the motion stays along one straight line in one direction — otherwise distance and displacement differ. Unless a direction is given, write "distance", not "displacement".
Name the gradient before calculating
Write "gradient = acceleration" (or "gradient = speed") before doing any arithmetic. This earns the method mark even if your tangent reading or numbers end up slightly outside the accepted range.
Use well-separated points for a gradient
When taking a gradient, pick two points as far apart as possible on the line or tangent. Wide spacing shrinks the percentage error in reading each coordinate and keeps your answer inside the accepted range.
Label axes with quantity and unit
Label each axis with the quantity AND its unit — "velocity / m/s" and "time / s", not just "v" and "t". Unlabelled axes lose the axis mark, and the unit also fixes the gradient unit (m/s ÷ s = m/s²) for you.
Distance = area: split it into shapes
For distance from a v-t graph, split the area into rectangles, triangles and trapezoids, work out each and add them. Name each shape as you go — a labelled area calculation is far easier to award marks to than a bare number.
| Quantity | Symbol | Formula | Unit |
|---|---|---|---|
| Speed (from d-t graph) | = gradient of d-t graph | m/s | |
| Acceleration (from v-t graph) | = gradient of v-t graph | m/s² | |
| Distance (from v-t graph) | = area under v-t graph | m | |
| SUVAT (uniform acceleration only) | — | ; ; | m/s, m/s², m |
Where = distance (m), = time (s), = velocity (m/s), = initial velocity (m/s), = acceleration (m/s²), = displacement/distance (m).
Important: SUVAT equations apply only when acceleration is constant (straight line on a v-t graph). For a curved v-t graph, use the area method.
What does the gradient of a distance–time graph represent?
A cyclist rides along a straight road. A distance–time graph of the journey shows a straight line from the point to the point .
(a) State what the gradient of a distance–time graph represents.
(b) Calculate the speed of the cyclist. Give the unit of your answer.
