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Physics graph skills

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

Practise

Plot a point

Plot P at time 6.5 s and position 3.5 cm. Tap the grid or enter both coordinates.

Plot the missing point PThe grid shows two existing points; P has not been plotted. Horizontal axis: Time / s, 0 to 8. Vertical axis: Position / cm, 0 to 10. Exact values are also listed below the graph.024680246810Time / sPosition / cmAB
Tap the grid to place P, or use the coordinate boxes below. Each small division is 0.5.
Larger graph
Plot the missing point PThe grid shows two existing points; P has not been plotted. Horizontal axis: Time / s, 0 to 8. Vertical axis: Position / cm, 0 to 10. Exact values are also listed below the graph.024680246810Time / sPosition / cmAB
Data to plot
Coordinates
PointTime / sPosition / cm
A12
B79
P6.53.5

Hint

Read time on the horizontal axis and position on the vertical axis. Move across first, then up.

Worked solution
  1. P is at (6.5 s, 3.5 cm).
Plot the missing point PThe grid shows two existing points; P has not been plotted. Horizontal axis: Time / s, 0 to 8. Vertical axis: Position / cm, 0 to 10. Exact values are also listed below the graph.024680246810Time / sPosition / cmAB
Another question
Question type: Mixed practice
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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