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Know what the axes mean before you reach for a formula. Practise gradients, areas, best-fit lines and the difference between a straight line and direct proportionality.
Original practice from Dr Desouky Physics Academy. The examples are teaching exercises, not exam-board questions.
gradient = Δy / Δx = (y₂ − y₁) / (x₂ − x₁)
Choose two well-separated points on a straight section or fitted line. Subtract in the same order for both axes, divide, then attach the vertical-axis unit divided by the horizontal-axis unit. Use coordinate changes, not the lengths of the lines measured with a ruler: changing the printed size must not change the Physics.
For the example below, (17 − 5) m ÷ (8 − 2) s = 2 m/s. Dividing 17 m by 8 s gives a different answer because this line does not pass through the origin.
Enter numbers in the displayed axis units. Decimal and e notation, such as 2.5e-3, are accepted. Convert milliseconds to seconds or milliamps to amps before entering them here.
Gradient · speed
2 m/s
Δy = 17 − (5) = 12 m.
Δx = 8 − (2) = 6 s.
Gradient = 12 ÷ 6 = 2 m/s.
The straight line through these points has a y-intercept of 1 m. Extending a fitted line to x = 0 is a mathematical extrapolation; whether it has a physical interpretation depends on the experiment.
For distance travelled against time, the gradient is speed. A decreasing total-distance graph would not describe a possible journey; use displacement to describe a change of direction.
Numerical results show up to 10 significant digits. This tool does not fit experimental data or calculate uncertainty. Round your final answer using the precision of the data and the question’s instructions.
The gradient of total distance travelled against time is speed. The gradient of displacement against time is velocity, which can be negative. A flat section means no change in distance or displacement during that interval. The height of a distance–time graph is a distance, not a speed; a steep line means a larger rate of change, not simply a larger distance.
The gradient is acceleration. The signed area between the graph and the time axis is displacement: count area below the axis as negative. Total distance is the sum of the magnitudes of those areas. For example, a velocity that rises uniformly from 2 m/s to 14 m/s over 6 s has acceleration 2 m/s² and displacement ½ × (2 + 14) × 6 = 48 m. These are different quantities with different units.
The area under a speed–time graph gives distance travelled. Speed is non-negative. A rectangle represents constant speed; a triangle or trapezium can represent a straight sloping segment. For a curve, estimate the area using a suitable grid or a series of narrow strips if the question requires an estimate.
For an ohmic resistor at constant temperature, plotting V vertically against I horizontally gives a line through the origin with gradient R. Swapping the axes gives gradient 1/R. For a non-ohmic device, resistance at a particular operating point is V/I; the gradient of a tangent is a different quantity. Reading the axis labels prevents a correct calculation from answering the wrong question.
A useful unit check: a velocity–time gradient has units (m/s)/s = m/s², while its area has units (m/s) × s = m. If your units give the wrong quantity, recheck the operation.
Draw a tangent matching the curve’s direction at the required point. Use two well-separated points on that tangent to find its gradient. A line connecting two points on the curve instead gives a secant gradient, an average rate over that interval. They answer different questions.
Sketch each situation and write the units before checking the answer. No graph-paper download or sign-up is needed.
Speed = (40 − 10) ÷ (8 − 2) = 5 m/s.
Use changes in both coordinates. The units are metres divided by seconds. Here the same section happens to extrapolate through the origin, but that is not a reason to skip the two-point method.
Acceleration = 2 m/s². Displacement = 32 m.
The gradient is (12 − 4) ÷ 4 = 2 m/s². The area is a trapezium: ½ × (4 + 12) × 4 = 32 m. Multiplying just the final velocity by the time would treat the velocity as constant.
No. Acceleration is −3 m/s², while speed increases from 3 m/s to 9 m/s.
Acceleration = [−9 − (−3)] ÷ 2 = −3 m/s². The velocity and acceleration have the same negative sign, so the magnitude of the velocity grows. A downward velocity–time slope does not always mean slowing down.
Displacement = +8 m. Total distance = 16 m.
The signed areas are +12 m and −4 m, giving +8 m. For total distance, add their magnitudes: 12 + 4 = 16 m. Equivalently, find the area under a speed–time graph.
The V–I gradient is 20 Ω. The I–V gradient is 0.050 A/V, or 0.050 S.
With V vertical, gradient = 6.0 ÷ 0.30 = 20 V/A. With I vertical, gradient = 0.30 ÷ 6.0 = 0.050 A/V, the conductance. Write down the axis quantities before deciding what a gradient represents.
No: it is linear, but it has a non-zero intercept.
Direct proportionality has the form y = kx and its line passes through the origin. For example, this line gives y = 5 at x = 1 and y = 8 at x = 2, so doubling x does not double y.
The Cambridge IGCSE Physics 0625 syllabus for 2026–2028 includes graph scales, gradients and intercepts; tangent gradients are identified as Extended content. The Edexcel International A-Level Physics specification includes graph analysis and experimental uncertainties. Check your own specification and the exact paper’s instructions for the assessed scope.
Identify the step you missed, try a fresh question, and return to it in a later session. Use the free Physics revision planner to make room for regular practice. For help applying these skills to your course, explore IGCSE Physics tutoring or A-Level Physics tutoring.
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