C.2 Algebra

Learning objectives

—C.2.1—Mathematical symbolsUnderstand and use =, <, ≪, ≫, >, ∝, ≈ and Δ in physical relationships.—C.2.2—Changing the subject of equationsChange the subject of equations, including non-linear equations such as rearranging E = mc².—C.2.3—Substitution into equationsSubstitute numerical values into algebraic equations using appropriate units for physical quantities.—C.2.4—Solving algebraic equationsSolve algebraic equations, including quadratic equations and kinematic equations for constant acceleration.

A mathematical symbol states how quantities are related. Reading the symbol precisely prevents an approximation, inequality or proportionality from being treated as an exact equality.

Symbol	Meaning in a physical relationship
$=$	is equal to
$<$ , $>$	is less than, is greater than
$\ll$ , $\gg$	is much less than, is much greater than
$\propto$	is proportional to; a constant of proportionality is not shown
$\approx$	is approximately equal to
$\Delta x$	change in $x$ , usually $x_{\mathrm{final}}-x_{\mathrm{initial}}$

$F\propto\frac{\Delta p}{\Delta t}$

This expression says that force increases with the rate of change of momentum. The two delta symbols define changes over the same interval; proportionality alone can be written as $F=k\Delta p/\Delta t$ only after introducing a constant $k$ .

Do not read $\propto$ as $=$ , $\approx$ as exact equality, or $\Delta p$ as the product $\Delta\times p$ . The signs of changes also depend on the chosen direction and on using final minus initial consistently.

The subject is the quantity isolated on one side of an equation. Change it by applying inverse operations to both sides, preserving equality at every line.

Current structure around the target	Inverse move
multiplied by a factor	divide both sides by that factor
divided by a factor	multiply both sides by that factor
raised to a power	apply the matching root
inside several operations	undo the outermost operation first

To make $m$ the subject of $E=mc^2$ , divide both sides by $c^2$ : $E/c^2=m$ , so $m=E/c^2$ . The square applies to $c$ only, so it remains with the whole denominator.

Check the rearrangement by substituting it back into the original equation: $mc^2=(E/c^2)c^2=E$ . A dimensional check also helps: $\mathrm{J}/(\mathrm{m^2\,s^{-2}})=\mathrm{kg}$ .

Moving a term across the equals sign is shorthand for performing the same operation on both sides. Do not change a sign or invert a factor without identifying that operation, and preserve brackets when a complete expression is squared or divided.

Substitution replaces each symbol by its measured numerical value and unit. Convert quantities to a compatible unit system first, then keep the equation's structure visible during the calculation.

Step	Action
1	write the equation and identify every symbol
2	convert prefixes and units, usually to coherent SI units
3	substitute values in brackets, especially negatives and powered terms
4	evaluate, simplify the unit and report justified precision

$p=mv$

For $m=0.450\,\mathrm{kg}$ and $v=12.0\,\mathrm{m\,s^{-1}}$ , $p=(0.450)(12.0)=5.40\,\mathrm{kg\,m\,s^{-1}}$ . If mass were supplied as $450\,\mathrm{g}$ , it would first become $0.450\,\mathrm{kg}$ .

Do not substitute a prefix as if it were part of the number: $5.8\,\mathrm{mm}=5.8\times10^{-3}\,\mathrm{m}$ , and a squared length carries the squared conversion factor. A unit attached only after unit-free arithmetic cannot expose an inconsistent substitution.

Solving a physics equation means finding every mathematical value of the unknown and then deciding which values satisfy the physical conditions. Kinematic equations such as $v=u+at$ and $s=ut+ frac12at^2$ apply only to constant acceleration.

Unknown's form	Useful route
appears once and linearly	isolate it with inverse operations
appears in a squared time term	collect terms into $At^2+Bt+C=0$
quadratic	factorise where possible or use $t=(-B\pm\sqrt{B^2-4AC})/(2A)$

If $s=20\,\mathrm{m}$ , $u=2.0\,\mathrm{m\,s^{-1}}$ and $a=3.0\,\mathrm{m\,s^{-2}}$ , then $20=2t+1.5t^2$ , so $1.5t^2+2t-20=0$ . The roots are about $3.05\,\mathrm{s}$ and $-4.38\,\mathrm{s}$ .

For elapsed time after the stated start, the positive root is the relevant solution, so report about $3.0\,\mathrm{s}$ at suitable precision. Substitution into the original equation checks both the arithmetic and the chosen root.

Do not discard a quadratic root merely because it is negative; reject or retain it using the defined origin, time interval and physical model. If acceleration is not constant, these kinematic equations do not apply over the interval.

A logarithm turns a ratio spanning many orders of magnitude into a compact scale value. On a base-10 logarithmic scale, adding 1 to the logarithm means multiplying the original ratio by 10.

$S=\log_{10}\left(\frac{Q}{Q_0}\right)$

The reference $Q_0$ makes the logarithm's argument dimensionless. If $Q/Q_0=10^3$ , then $S=3$ ; if the ratio becomes $10^5$ , the scale rises by 2 even though the original quantity is multiplied by $10^2=100$ .

Sound intensity level is a real-world example: $L=10\log_{10}(I/I_0)\,\mathrm{dB}$ . An intensity ratio of $100$ gives $L=10\log_{10}(100)=20\,\mathrm{dB}$ relative to $I_0$ .

This logarithmic-scale skill is full A Level content. Equal steps on the scale represent equal multiplying factors, not equal additions to the original quantity; never take the logarithm of a dimensional value without first forming the defined ratio.

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—C.3 GraphsRead

C.2 Algebra

Learning objectives

Read every mathematical symbol as a relationship

Change the subject with reversible operations

Substitute values only after aligning their units

Solve the equation, then test the physical solution

A logarithmic scale records multiplicative change

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