Home > Subjects > Physics > Level 2 > 2.1 Physics investigation > Key tips

- Subject: Physics
- AS: 91168
- Level: 2
- Credits: 4
- Internal

## Physics 2.1 Carry out a practical physics investigation that leads to a non-linear mathematical relationship

### Key tips

Follow the written instructions that will be given to you by your teacher.

Ask lots of questions.

If you draw a raw-data graph, make sure you can recognise from the shape of the graph the four types of relationships (linear, square, inverse, and inverse square).

You will be provided with a formula for the relationship between the variables. Use this to help decide the type of relationship and which transformations will give you a straight line graph. For example if you are investigating the relationship between the distance travelled by an accelerating object, and the time since it was at rest, the relevant formula is d= ½at^{2} . So a graph of ‘d’ against ‘t^{2}’ will produce a straight line with a gradient of ½a.

Make sure you prepare for the assessment by knowing how to transform data from a non-linear relationship to give a straight-line graph.

Draw a careful and accurate straight-line graph with both axes carefully labelled with quantity and unit. Don’t forget to transform your units. For example, if one of your graphed variables is time squared (t^{2}) then the units for that axis will seconds squared (s^{2}). The axis label will look like this: t^{2}(s^{2}).

Draw a Line of best fit (LOBF) and use it to work out the gradient. Be careful to use the line, not your data table, to find the gradient. Show how you worked out the gradient by drawing a triangle on the LOBF.

The units of the gradient are the units of the ‘y’ axis over the units of the ‘x’ axis. For example, a graph of distance against time-squared will have units m/s^{2} or ms^{-2}.

Give the mathematical relationship in terms of the actual variables, not ‘y’ and ‘x’. For example a graph of ‘d’ against ‘t^{2}’ with a gradient of 5.1 gives a relationship d=5.1t^{2}.

You can use your relationship to work out a value for a physical quantity or constant. For example a relationship of d=5.1t^{2} can be equated with the provided formula d= ½at^{2} to give a=10.2ms^{-1}.