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Linearising relationships to y = mx + c

Edexcel IAL Physics · WPH13 Practical Skills in Physics I

Straight lines are easy to test and easy to measure a gradient from. Choosing what to plot so the theory predicts a straight line is a core Unit 3 skill.

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The idea

Rearrange the equation so it matches y = mx + c, with measured quantities (or simple functions of them) as y and x, and constants in m and c:

Relationship	Plot y	against x	Gradient	Intercept
h = (1/2)gt² (CP1)	h	t²	g/2	0
T = 2π sqrt(l/g)	T²	l	4π²/g	0
R = ρl/A (CP7)	R	l	ρ/A	0 (contact resistance if not)
V = -rI + ε (CP8)	V	I	-r	ε
f = (1/2L) sqrt(T/μ) (CP5)	f	1/L	(1/2) sqrt(T/μ)	0
I = k/d²	I	1/d²	k	0
v² = u² + 2as	v²	s	2a	u²

Showing that a graph tests a relationship

Examiners want three things:

the equation rearranged and compared explicitly with y = mx + c (write the + and = signs; arrows and loops are not credited);
a statement that the gradient equals a constant (name it);
if the comparison is with y = mx + c, a statement that the intercept is zero (or what it equals).

The relationship is confirmed if the graph is a straight line (through the origin, where predicted).

Processing the data

Add a column for the new quantity (t², 1/L, 1/d²) and give it to the same number of significant figures as the raw data. A classic slip is misreading 1/d² as I/d²: read the question carefully.

Worked example

A graph of T² against l for a pendulum has gradient 4.02 s²/m. Find g.

T² = (4π²/g) l, compared with y = mx: gradient = 4π²/g
g = 4π² / 4.02 = 9.821

Answer: g = 9.82 m/s²

Key idea: Rearrange into y = mx + c with constants in m and c. State that the gradient is a constant (named) and the intercept is zero where predicted; a straight line confirms the relationship.

Check yourself

A graph of T² against l has gradient 4.05 s²/m. Find g in m/s², to 3 s.f.
Show answer
9.75 m/s². g = 4π² / 4.05 = 9.75 m/s².
In core practical 1, a graph of h against t² has gradient 4.90 m/s². Find g in m/s².
Show answer
9.80 m/s². Gradient = g/2, so g = 2 x 4.90 = 9.80 m/s².
To find a constant acceleration from v² = u² + 2as, which graph should be plotted?
- v against s, gradient a
- v² against s, gradient 2a
- v against s², gradient 2a
- s against v, gradient 1/2a
Show answer
v² against s, gradient 2a. v² = 2a s + u² matches y = mx + c with y = v² and x = s.
Intensity is thought to follow I = k/d². Which graph should be a straight line through the origin?
- I against d²
- I against d
- I against 1/d²
- 1/I against d
Show answer
I against 1/d². I = k x (1/d²) is y = mx with x = 1/d².
Calculate 1/d² for d = 0.250 m, in m^-2, to 3 s.f.
Show answer
16.0 m^-2. 1 / 0.0625 = 16.0 m^-2.
In core practical 5, f is plotted against 1/L at fixed tension and mass per unit length. What does the gradient equal?
- sqrt(T/μ)
- 2L
- (1/2) sqrt(T/μ)
- T/μ
Show answer
(1/2) sqrt(T/μ). f = (1/2) sqrt(T/μ) x (1/L).
Explain how a graph of T² against l would confirm that T = 2π sqrt(l/g).
Show answer
Squaring gives T² = (4π²/g) l, which is of the form y = mx with gradient 4π²/g, a constant. If the graph is a straight line through the origin, the relationship is confirmed.. Compare with y = mx, name the constant gradient, and state the shape expected.

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Last updated 8 November 2026. Spotted a mistake? Tell us.