Physics graph skills

Read scales, plot a point and use gradients and intercepts to interpret Physics graphs.

Practise

Choose a best-fit line

Choose the line that best represents the trend. Calculate its gradient in cm s⁻¹. Give your answer to 3 significant figures.

Read the axes and units. The values are also available in the table below.

Larger graph

Graph values

Synthetic readings
Point	Time / s	Position / cm
1	1	4.2
2	2	4.6
3	3	6.2
4	4	6.8
5	5	8.4
6	6	8.8

Line A: (0, 1.8) to (7, 8.8).
Line B: (0, 3) to (7, 10).
Line C: (0, 4.2) to (7, 11.2).

Treat the displayed values as exact unless an uncertainty is stated.

Hint

A best-fit line need not pass through every point. Use two well-separated positions on the line to calculate Δy / Δx.

Worked solution

Selection: Line B.
Line B balances the points above and below; the other lines are displaced vertically.
Two positions on line B: (0 s, 3 cm) and (7 s, 10 cm).
Gradient = (10 − 3) / (7 − 0) = 1 cm s⁻¹.
Answer: 1.00 cm s⁻¹ (3 significant figures).

Another question

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.

Find uncertainty in a gradient

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