Physics: Graph Drawing

    This applies to both AS and A Level exams, with the exception of the blue subheadings, which are A Level only.
    You can find more examples in Appendix 10 in the Edexcel specification.

What to plot:

  • You'll need to rearrange to the form y=mx+cy = mx + c where your two variables are yy and xx
  • 'A graph of aa against bb' means that the first variable mentioned goes on the yy-axis and the second goes on the xx-axis
  • 'directly proportional to' can be written as =k=k where kk is a constant
  • For example, to test if the measurement aa is directly proportional to bb, we need to find if it satisfies the equation a=kba = kb
  • This can be rearranged to y=mx+cy = mx + c form: a=kb+ca = kb + c
  • Therefore, we need to plot aa against bb. It should form a straight line with a yy-intercept of 0 (because c=0c = 0) and gradient kk
  • Use this method to determine whether measurements satisfy a relationship
  • 'inversely proportional' means a=k1xa = k\frac{1}{x}


  • Axes must be labelled with the quantity that it represents, and the units
  • For example, if you're plotting distance squared on the xx-axis, you could label it 'd2 (m2)d^2\ (\mathrm{m}^2)'
  • Although writing d2/m2d^2 /\mathrm{m}^2' is preferred by SI

Units in log axes:

  • If you're plotting the log of distance, you would write log(d/m)\log(d/\mathrm{m})'

Error bars:

  • Usually, you would pick a measurement in the middle, calculate the absolute error in that value, and use that for every error bar

Example 1:

  • Let's say you're taking measurements of the time of something
  • Pick the middle reading
  • Let's say the measurements of time for the middle reading are 1.42 s1.42\ \mathrm{s}, 1.46 s1.46\ \mathrm{s}, 1.67 s1.67\ \mathrm{s}, 1.49 s1.49\ \mathrm{s} and 1.43 s1.43\ \mathrm{s}
  • 1.67 s1.67\ \mathrm{s} should be discarded because it's an anomaly
  • This gives a mean of 1.45 s1.45\ \mathrm{s}. Mark this as a data point
  • Our minimum value is 1.42 s1.42\ \mathrm{s} (0.03 s0.03\ \mathrm{s} away from the mean) and maximum is 1.49 s1.49\ \mathrm{s} (0.04 s0.04\ \mathrm{s} away from the mean)
  • Therefore, pick the largest distance from the mean (0.04 s0.04\ \mathrm{s}). This will be the length of our error bar in each direction
  • So draw a bar of width 0.04×2=0.04 \times 2 = 0.08 s0.08\ \mathrm{s} on each data point

Example 2:

  • We'll use the same values as above
  • Except this time, we need to plot ln(t/s)\ln(t/\mathrm{s})
  • The mean value is 1.45 s1.45\ \mathrm{s}, so plot the natural log of this (0.3720.372)
  • The largest distance from the mean is 0.04 s0.04\ \mathrm{s}
  • This makes the maximum possible value 1.45+0.04=1.491.45 + 0.04 = 1.49 s\mathrm{s} and the minimum 1.450.04=1.411.45 - 0.04 = 1.41 s\mathrm{s}
  • The length of the error bar is therefore ln(1.49)ln(1.41)=0.0552\ln(1.49) - \ln(1.41) = 0.0552

Line of best fit:

  • First, draw the best fit line through the data points
  • Now draw the error boxes and draw the steepest and shallowest possible line which passes through all of them
  • Points which are clearly anomolous can be ignored when drawing a line of best fit