Units, prefixes and standard form
Convert prefixes, squared and cubed units. Write numbers in standard form.
Physics graph skills
Read scales, plot a point and use gradients and intercepts to interpret Physics graphs.
Practise
Choose axesA resistor obeys V = IR. Choose axes so the gradient gives resistance directly.
Data to plot
| Point | Current / A | Potential difference / V |
|---|---|---|
| 1 | 0.02 | 0.4 |
| 2 | 0.04 | 0.8 |
| 3 | 0.06 | 1.2 |
| 4 | 0.08 | 1.6 |
Hint
Compare V = IR with y = mx. The coefficient of the horizontal quantity is the gradient.
Worked solution
- Selection: Horizontal: current / A; vertical: potential difference / V.
- Plot potential difference vertically against current horizontally.
- Gradient = ΔV / ΔI, with units V A⁻¹ = Ω.
Question type: Mixed practice
Report a problem
Calculate a graph gradient from two points
A potential-difference–current line passes through (0.10 A, 1.5 V) and (0.50 A, 7.5 V). Gradient = (7.5 − 1.5)/(0.50 − 0.10) = 15 V A⁻¹ = 15 Ω. Choose separated points on the fitted line; they need not be measured data points.
Read which quantity is on each axis. If current is vertical and potential difference horizontal, the gradient is conductance, 1/R. Dividing one coordinate by the other gives the gradient only when the relevant straight line passes through the origin.
Read the scale before calculating
Count the small divisions between two labelled values. If 10 divisions span 2 seconds, each division is 0.2 s. Check whether the scale starts at zero.
Gradient and intercept
For a straight line, gradient = Δy / Δx. Use two well-separated points on the line and subtract in the same order. A falling line has a negative gradient. Use changes in the coordinates; y / x only gives the gradient when the line passes through the origin.
The vertical intercept is the value of y when x = 0. On a displacement–time graph it gives initial displacement, while the gradient gives velocity in m s⁻¹.
Best-fit lines
Follow the overall trend and balance points above and below the line. A best-fit line need not pass through a data point. The readings here are synthetic practice data; the exercise asks you to choose between three candidate lines.
Use the axes to interpret the gradient
For V against I, the gradient is V / I = R, in Ω. Reversing the axes gives 1 / R. For extension x against force F, x = F / k: the gradient is 1 / k. Convert extension to metres before finding k in N m⁻¹.
What this practice includes
Linear scales, plotting one point, signed gradients, intercepts, choosing a straight best-fit line, axes and two physical interpretations. Displayed values are treated as exact for practice. Numerical answers use 3 significant figures; plotting coordinates must match the stated values exactly.
Curve fitting, tangents, areas, logarithmic axes and gradient uncertainty are not included yet.
A-level relevance
AQA 7408: MS 3.2–3.5 (partial); AQA 7408: 3.4.1.3, 3.4.2.1 (selected); OCR A H556: 1.1.3(d)(i–ii), 3.1.1(c), 3.4.1(b–d) (partial). Tasks cover the selected skills listed above.
AQA specification · AQA mechanics and materials specification · OCR A specification