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⚡ Pure Physics 6091 📈 Paper 2 Essential Skill

Physics Graph Skills

Graph questions appear in every O-Level Physics Paper 2. This page covers every skill you need: drawing best-fit lines, calculating gradients, drawing tangents to curves, reading intercepts, identifying anomalous results, and interpreting what each graph shape means physically.

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Skill 1 — Drawing Lines of Best Fit
Straight lines and smooth curves through experimental data
Annotated graph showing how to calculate gradient using delta y over delta x and area under line Reading a Straight-Line Graph x-axis (e.g. time / s) y-axis y-intercept (c) Delta x Delta y gradient = Dy / Dx Area = distance travelled y = mx + c m = gradient, c = y-intercept
Graph Skills — gradient, area and y-intercept on a straight-line graph

What is a line of best fit?

A line of best fit (or curve of best fit) is a single line or smooth curve that passes as close as possible to all data points, with approximately equal numbers of points above and below it. It represents the underlying trend in the data, smoothing out random experimental errors.

A line of best fit does NOT have to pass through the origin unless there is a physical reason it should (e.g. V=IR passes through origin if resistance is constant and y-intercept is zero). A line of best fit also does NOT have to pass through any specific data point — it balances the scatter on both sides.

Rules for drawing best-fit lines

  • 1
    Plot all points first — use small, neat crosses (×) or dots with circles (⊙). Never use thick blobs.
  • 2
    Identify anomalous points — any point clearly far from the trend should be circled and excluded from the line. Do not include anomalies in best-fit line.
  • 3
    Straight line or curve? — If the pattern is linear, use a ruler. If curved, draw a single smooth freehand curve. Never draw a dot-to-dot jagged line.
  • 4
    Balance the points — roughly equal numbers of points above and below the line/curve along its full length.
  • 5
    Extend the line — if asked for the y-intercept or x-intercept, extend the line to the relevant axis using a dashed line extension.
⚠ Most common errors

Drawing a dot-to-dot line through every point instead of a smooth best-fit. Forcing the line through the origin when it should not be. Including anomalous points in the best-fit. Drawing a thick line — use a sharp pencil.

x-axis (independent variable) y-axis (dependent variable) 0 1 2 3 4 5 anomalous best-fit line (not dot-to-dot)
Correct best-fit line — balances points above and below. Anomalous point circled and excluded. Line extended to axes.
Q
A student investigates how the extension of a spring depends on the load applied. Results: Load (N): 0, 1, 2, 3, 4, 5. Extension (cm): 0, 2.1, 4.0, 5.8, 9.5, 9.8. Identify which reading is anomalous and explain how you would draw the best-fit line for the remaining data. [3]
▶ Show answer
Answer: The reading at 4 N (9.5 cm) is anomalous — it is far above the trend suggested by the other readings and does not fit the linear pattern. [1] Circle it and exclude it from the best-fit line. [1] Draw a straight best-fit line through the remaining five points (0,0), (1,2.1), (2,4.0), (3,5.8), (5,9.8), with approximately equal numbers of points above and below the line, using a ruler. The line passes close to or through the origin since no load = no extension. [1]
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Skill 2 — Calculating the Gradient
Rise over run — using two points far apart on the best-fit line
Gradient formula
gradient = Δy / Δx = (y₂ − y₁) / (x₂ − x₁)
Use two points FAR APART on the best-fit line — not data points. The larger the triangle, the more accurate the gradient.

Step-by-step gradient calculation

  • 1
    Choose two points on the LINE — not data points, unless they happen to sit exactly on the line. Use points as far apart as possible.
  • 2
    Draw a large right-angle triangle — mark it clearly on the graph with dotted lines.
  • 3
    Read off Δy and Δx — using the axis scale, read the vertical and horizontal spans of your triangle.
  • 4
    Calculate: gradient = Δy / Δx — include correct units (y-units / x-units).
  • 5
    Check the sign — positive gradient means y increases as x increases. Negative gradient means y decreases as x increases.
✓ Rule: gradient has physical meaning

The gradient of a v-t graph = acceleration (m/s²). The gradient of a distance-time graph = speed (m/s). The gradient of a V-I graph = resistance (Ω). Always state what the gradient represents and include units.

⚠ Common errors

Using data points instead of points on the line. Using a small triangle (amplifies reading errors). Forgetting units. Not using the full axis scale when reading coordinates.

v-t graph gradient

Gradient = Δv/Δt = change in velocity / time = acceleration. Units: (m/s)/s = m/s². A negative gradient = deceleration.

d-t graph gradient

Gradient = Δd/Δt = distance / time = speed. Units: m/s. Steeper gradient = faster speed. Horizontal section = stationary.

V-I graph gradient

Gradient = ΔV/ΔI = resistance R. Units: V/A = Ω. For an ohmic conductor the graph is a straight line through origin — constant gradient = constant resistance.

Force-extension graph

Gradient = ΔF/Δx = spring constant k. Units: N/m or N/cm. The gradient gives the stiffness of the spring.

Q
A velocity-time graph has two points on the best-fit line: (2 s, 8 m/s) and (10 s, 32 m/s). Calculate the gradient and state what it represents. [3]
▶ Show answer
Answer: Gradient = (y₂−y₁)/(x₂−x₁) = (32−8)/(10−2) = 24/8 = 3 m/s² [1 formula, 1 calculation]. The gradient represents the acceleration of the object — the rate of change of velocity with time. [1 for correct physical interpretation with unit]
Q
A student plots a current (y-axis, A) vs voltage (x-axis, V) graph. Two points on the best-fit line are (1.0 V, 0.20 A) and (5.0 V, 1.00 A). Calculate the gradient, state its unit, and explain what it represents. [3]
▶ Show answer
Answer: Gradient = ΔI/ΔV = (1.00−0.20)/(5.0−1.0) = 0.80/4.0 = 0.20 A/V [1 calculation, 1 unit]. The gradient represents the reciprocal of resistance (1/R = conductance). R = 1/gradient = 1/0.20 = 5 Ω. [1 for correct physical interpretation] Note: if the graph were V (y-axis) vs I (x-axis), the gradient would be R directly.
✏️
Skill 3 — Drawing Tangents to Curves
Finding the rate of change at a specific point on a curved graph

Why tangents are needed

When a graph is curved (not straight), the gradient is not constant — it changes at every point. To find the rate of change at a specific point, you draw a tangent: a straight line that just touches the curve at that point without crossing it (except at the point itself).

The gradient of the tangent at a point = the instantaneous rate of change of the variable at that point. This is used to find: instantaneous acceleration from a d-t graph; instantaneous rate of reaction from a volume-time curve; instantaneous speed from a position-time curve.

How to draw a tangent correctly

  • 1
    Mark the point on the curve where you need the gradient.
  • 2
    Place a ruler so it touches the curve at that point only — the ruler should not cut through the curve on either side of the point.
  • 3
    Adjust the angle — rotate the ruler until the gaps between the curve and the ruler are equal on both sides of the point. The curve should "mirror" away from the ruler symmetrically.
  • 4
    Draw a long straight line — the longer the tangent line, the easier to calculate a precise gradient. Extend it well beyond the point in both directions.
  • 5
    Calculate the gradient of the tangent using Δy/Δx with a large triangle, as in Skill 2.
⚠ Common errors

Drawing a chord (line connecting two points on the curve) instead of a tangent. Drawing the tangent correctly but then using a tiny triangle to calculate gradient. Not labelling which point the tangent is drawn at.

Q
A distance-time graph shows a curve (not a straight line). Explain how you would find the speed of the object at exactly t = 4 s. [3]
▶ Show answer
Answer: Draw a tangent to the distance-time curve at the point where t = 4 s — a straight line that just touches the curve at that point without crossing it nearby. [1] Calculate the gradient of this tangent by drawing a large right-angle triangle and reading off Δd and Δt from the axes. [1] Gradient = Δd/Δt = speed at t = 4 s (units: m/s). [1]
Q
In a rate-of-reaction experiment, a student plots volume of gas collected (cm³) against time (s). The graph is a curve that levels off. The tangent at t = 20 s passes through (5 s, 10 cm³) and (45 s, 90 cm³). Calculate the rate of reaction at t = 20 s. [2]
▶ Show answer
Answer: Rate = gradient of tangent = Δy/Δx = (90−10)/(45−5) = 80/40 = 2.0 cm³/s [1 method, 1 answer with unit]. This is the instantaneous rate of gas production at t = 20 s. The rate at the start (t = 0) would be steeper (larger gradient); as the reaction slows, the tangent gradient decreases.
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Skill 4 — Interpreting Graph Shapes
What each curve shape means physically in O-Level Physics

Straight line through origin

Meaning: Direct proportion. y ∝ x. e.g. V vs I (ohmic resistor), F vs a (Newton's 2nd Law for constant mass), extension vs force (Hooke's Law within limit).

Straight line, y-intercept ≠ 0

Meaning: Linear relationship but NOT direct proportion. e.g. Temperature vs time for a heating substance; velocity vs time with initial velocity (SUVAT: v = u + at, intercept = u).

Curve, increasing gradient

Meaning: Rate of change is increasing. e.g. Distance vs time for an accelerating object (d = ½at²); exponential growth. As x increases, y increases faster and faster.

Curve, decreasing gradient (levelling off)

Meaning: Rate of change is decreasing. e.g. Rate of reaction slowing as reactants consumed; terminal velocity being approached; radioactive decay curve. As x increases, y increases more and more slowly.

Horizontal line

Meaning: y is constant, independent of x. e.g. Constant velocity on v-t graph (zero acceleration); temperature during state change (latent heat); terminal velocity plateau.

Exponential decay curve

Meaning: y decreases by the same fraction in equal intervals of x. e.g. Radioactive decay (activity vs time). The curve approaches the x-axis but never reaches it. Half-life is read from the graph.

Q
A student plots current (y-axis) vs potential difference (x-axis) for a filament bulb. Describe and explain the expected shape of the graph. [3]
▶ Show answer
Answer: The graph curves — it is not a straight line. [1] At low voltages, the current increases approximately proportionally (nearly straight). At higher voltages, the curve bends — the gradient decreases (current increases more slowly than voltage). [1] This is because as current increases, the filament heats up significantly. Higher temperature increases the resistance of the metal filament (more collisions between electrons and vibrating metal ions). By V=IR, higher R at the same V means less current than a straight-line graph would predict. [1]
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Skill 5 — Anomalies, Intercepts & Extrapolation
Reading beyond data range and identifying outliers

Anomalous results

An anomalous (outlier) result is a data point that does not follow the general trend and lies clearly away from the best-fit line. To deal with anomalous results: (1) circle the point, (2) exclude it from the best-fit line, (3) repeat the measurement if possible, (4) comment on a possible cause (e.g. timing error, contamination, misreading the scale).

⚠ Never say "human error"

Always specify the exact source: "parallax error when reading the meniscus in the burette" or "reaction time error when starting the stopwatch at the moment of mixing." "Human error" without specification earns no marks.

Reading intercepts

y-intercept: the value of y when x = 0. Extend the best-fit line to the y-axis if necessary. In v = u + at, the y-intercept = u (initial velocity). In y = mx + c, y-intercept = c.

x-intercept: the value of x when y = 0. Extend the line to the x-axis. Less commonly tested but may appear in cooling curves or charging/discharging problems.

Extrapolation: extending the best-fit line beyond the data range to estimate values. This assumes the same relationship continues — state this assumption. Extrapolation is less reliable than interpolation (reading within the data range).

Q
A velocity-time graph for a uniformly accelerating object has a y-intercept of 6 m/s and passes through (8 s, 30 m/s). Find (a) the initial velocity, (b) the acceleration, (c) the time at which the object was at rest (if it was decelerating before this). [4]
▶ Show answer
Answer:
(a) Initial velocity = y-intercept = 6 m/s. [1]
(b) Gradient = (30−6)/(8−0) = 24/8 = 3 m/s² (acceleration). [1]
(c) The object is moving in the positive direction throughout (velocity goes from 6 m/s upward, never crosses zero in this scenario). If instead the question described initial velocity as negative (decelerating from rest), the x-intercept gives the time at which v=0. Using v=0, 0=6+3t is not applicable here since gradient is positive. If u were −6 m/s with gradient +3: 0=−6+3t → t = 2 s. [1 for method, 1 for answer in a suitably re-worded version]

📋 Quick Reference — Graph Skills Checklist

✓ Plot points as neat crosses (×)
✓ Circle and exclude anomalous points
✓ Draw line/curve — never dot-to-dot
✓ Use large triangle for gradient
✓ Use points ON the line, not data
✓ Include units on gradient answer
✓ Tangent: touch curve at one point only
✓ State what gradient represents physically
✓ Label axes with quantity AND unit
✓ Scale must use at least 50% of grid

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