Change the subject of the formula $v=u+at$ to $u$.
Transposition or changing the subject of a formula
This topic can be seen as an extension of the work we have done on solving linear equations and gives us the tools we need to work with formulas of all kinds.
In this section, we will look at various techniques for taking a formula and manipulating it so that we can use it to find other variables.
For example, here is a very simple and very profound formula: $F=ma$. This formula is one of the mainstays of physics and is the core principle behind Newton's second law of motion. This formula allows us to find the Force acting on a body when the mass of the body is $m$ and its acceleration is $a$.
Transposition will allow us to create formulas, where given the values of $m$ and $F$, we can find $a$ or if we are given the values of $F$ and $a$, we can find the value of $m$.
Worked example 1: Transposition of formulas #1 (One step)
 Notes
Notation & language
The subject of a formula is the letter on the LHS of the equal sign. This is not always the case, as there are some more advanced situations like $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, but in the vast majority of cases you will encounter at GCSE (16 year olds), it will be a single letter.
In this example, $v$ is the subject of the formula.
You may also come across the word transposition or transpose, which means the same as change the subject of the formula.
Flowchart method
\[\begin{align} \therefore \ u=vat \\ \hline \end{align} \]
Balancing method
\[\begin{align} v&=u+at \\ vat&=u+atat \\ vat &= u \\ \therefore \ u&=vat\\ \hline \end{align}\]
Flow chart method
We will use the principles that we have encountered in solving equations to perform
In this first example, I shall return to the flowchart method for solving equations, so you can see the underlying concepts here, but subsequent formula transposition will be solved using more standard balancing techniques.
 Start by writing down the variable we want to make the subject, ($u$).
 Examine the RHS of the formula. What happens to $u$ as the formula is built? This tells us that $+at$ is the first (and only) operation.
 We now put the current subject, which is equal to the RHS we have built up, into the machine backwards.
 The inverse of $+at$ is $at$ and this operation gives us $vat$ which is equal to $u$
Balancing method
 Look at the original formula, allow your attention to rest upon the variable you want to make the subject of the formula.
 What is done to the $u$ to build the RHS? $+at$.
 The inverse of this is $at$, so we must subtract $at$ from both sides of the formula.
 This gives us the answer.
 \[u=vat\]
Worked example 2: Transposition of formulas #2 (Two steps)
 Notes
Change the subject of the formula $v=u+at$ to make $t$ the subject.
 \[\begin{align} v&=u+at \\ vu&=u+atu \\ vu&=at \\ \frac{vu}{a}&=\frac{at}{a} \\ \frac{vu}{a}&=t \\ t&=\frac{vu}{a} \\ \end{align}\]
This problem is rather more awkward than the previous example, despite it being the same formula. This is because the $t$ is rather more deeply embedded in the formula than $u$ is.
 Start, as before by examining the RHS is reference to the new subject, in this case $t$.
 There are 2 steps this time. The first is $\times a$, the second is $+u$.
 Doing the inverses from the end, we do $u$ from both sides.
 And then we must $\div a$, giving us the correct answer.
It is not strictly necessary to swap the LHS & RHS, as I have done here, but I prefer it so I have.
 \[t=\frac{vu}{a}\]
Worked example 3: Transposition of formulas #3 (Selfinverse problems)
 Notes
 Change the subject of the formula $a = 3  b$ to make $b$ the subject.
 Make $q$ the subject of the formula $p=\frac{5}{q}$.
Look carefully at these two problems. These are examples of selfinverse operations.
Q: What happens to $b$? A: it is subtracted from $3$. This is wholly different to subtract $3$, whose inverse is $+3$.
The inverse of subtract from 3 is subtract from 3. Take the number $2$. Subtract it from $3$ and you get $1$. The inverse is the operation which gets us back to $2$ again.
We do this by subtracting from 3 again. i.e. $31=2$
 a.
Flowchart method
Balancing method
\[\begin{align} b&=3a\\ b+a&=3a+a\\ b+a&=3\\ b+ab&=3b\\ a&=3b\\ \hline \end{align}\]
b.Flowchart method
Balancing method
\[\begin{align} p &= \frac{5}{q} \\ pq &= \frac{5}{q} \times q \\ pq &= 5\\ \frac{pq}{p}&=\frac{5}{p} \\ q&=\frac{5}{p} \\ \hline \end{align}\]The flow chart method is interesting here, because of the selfinverse nature of the operation
The balancing method shows that going the long way round gives the same answer, but noticing the self inverse nature of both Subtract from 3 and in part b. Divide into 5. This can save you some time if you understand how this works.
If it makes no sense to you at this stage, move on. You can solve these in other ways and this is nonessential and not worth hurting your head over.
 $a=3b$
 $q=\frac{5}{p}$
Worked example 4: Transposition of formulas #4 (Hard ones with more than one instance of the subject variable)
 Notes
Make $d$ the subject of the formula
\[c=\frac{d+1}{d3}\]This is the first instance we have come across where this cannot be solved using the flowchart method because of the fact that there are $d$s in both the numerator and denominator of the fraction.
The difficulty here in how to get from having two $d$s in the formula to having just one.
 \[\begin{align} c&=\frac{d+1}{d3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ c(d3)&=\frac{d+1}{d3} \times (d3) \ \ \ \ (2) \\ c(d3)&=d+1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ cd3c&=d+1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \\ cd3c+3c&=d+1+3c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\ cd&=d+1+3c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) \\ cdd&=d+1+3cd \ \ \ \ \ \ \ \ (7) \\ cdd&=1+3c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8) \\ d(c1)&=1+3c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9) \\ \frac{d(c1)}{c1}&=\frac{1+3c}{c1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10) \\ d&= \frac{1+3c}{c1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11) \\ \hline \end{align}\]
First off, these are a bit tricky and should not be attempted unless you are really familiar with the work which has lead up to this point.
 Write down the formula (I know I keep saying this, but it is truly remarkable how many students appear to think that this does not apply to them.)
 Start by getting rid of the fraction. So we $\times (d3)$
 Tidy up.
 Multiply out the brackets on the LHS.
 Add $3c$ to both sides.
 Tidy up.
 Subtract $d$ from both sides.
 Tidy up.
 Factorise the LHS, taking out the $d$ common to both terms.
 Divide both sides by $(c1)$.
 Tidy up.
 \[d= \frac{1+3c}{c1}\]
Exercise 1: Transposition of formulas #1 (One step problems)
 Instructions:
In the following problems, make the letter in the the bracket, the subject of the formula.
 Q1\[h=ga \ \ \ \ \ (g)\]\[g=a+h\]
 Q2\[X=Y+Z \ \ \ \ \ (Y)\]\[Y=XZ\]
 Q3\[e=f7 \ \ \ \ \ (f)\]\[f=e+7\]
 Q4\[P=QR \ \ \ \ \ (Q)\]\[Q=P+R\]
 Q5\[v=u+4t \ \ \ \ \ (u)\]\[u=vat\]
 Q6\[s=y3t \ \ \ \ \ (y)\]\[y=s+3t\]
 Q7\[T=2S+P \ \ \ \ \ (P)\]\[P=T2S\]
 Q8\[b=d+3c \ \ \ \ \ (d)\]\[d=b3c\]
 Q9\[g=mpq \ \ \ \ \ (m)\]\[m=g+pq\]
 Q10\[u=v7w \ \ \ \ \ (v)\]\[v=u+7w\]
 Q11\[c=x+v \ \ \ \ \ (x)\]\[x=cv\]
 Q12\[y=xqy \ \ \ \ \ (x)\]\[x=y+qy\]
 Q13\[c=2d \ \ \ \ \ (d)\]\[d=\frac{c}{2}\]
 Q14\[d=5e \ \ \ \ \ (e)\]\[e=\frac{d}{5}\]
 Q15\[g=ha \ \ \ \ \ (a)\]\[a=\frac{g}{h}\]
 Q16\[r=st \ \ \ \ \ (t)\]\[t=\frac{r}{s}\]
 Q17\[y=\frac{1}{2}x \ \ \ \ \ (x)\]\[x=2y\]
 Q18\[t=\frac{s}{3} \ \ \ \ \ (s)\]\[s=3t\]
 Q19\[i=\frac{j}{k} \ \ \ \ \ (j)\]\[j=ik\]
 Q20\[m=\frac{n}{2p} \ \ \ \ \ (n)\]\[n=2mp\]
Exercise 2: Transposition of formulas #2 (Two step problems)
 Instructions:
In the following problems, make the letter in the the bracket, the subject of the formula.
As you proceed, try to spot patterns between problems, looking for similarities and differences.
 Q1\[a=2b+c \ \ \ \ \ (b)\]\[b=\frac{ac}{2}\]
 Q2\[m=np4 \ \ \ \ \ (p)\]\[p=\frac{m+4}{n}\]
 Q3\[v=u+at \ \ \ \ \ (a)\]\[a=\frac{vu}{t}\]
 Q4\[v=u+at \ \ \ \ \ (t)\]\[t=\frac{vu}{a}\]
 Q5\[a=2d+e \ \ \ \ \ (d)\]\[d=\frac{ae}{2}\]
 Q6\[x=3yz \ \ \ \ \ (y)\]\[y=\frac{x+z}{3}\]
 Q7\[x=3y+4z \ \ \ \ \ (y)\]\[y=\frac{x4z}{3}\]
 Q8\[q=p^2+3r \ \ \ \ \ (r)\]\[r=\frac{qp^2}{3}\]
 Q9\[d=4e^2+3f \ \ \ \ \ (e)\]\[e=\sqrt{\frac{d3f}{4}}\]
 Q10\[t=\sqrt{u4}+v \ \ \ \ \ (u)\]\[u=(tv)^2+4\]
 Q11\[C=D(EF) \ \ \ \ \ (E)\]\[E=F + \frac{C}{D}\]
 Q12\[G=2H(2I+J) \ \ \ \ \ (I)\]\[I=\frac{G}{4H}  \frac{J}{2}\]
 Q13\[f=\frac{gh}{i} \ \ \ \ \ (g)\]\[g=fi+h\]
 Q14\[l=\frac{2m+n}{p} \ \ \ \ \ (m)\]\[m=\frac{lpn}{2}\]
 Q15\[m=\frac{2np}{2q} \ \ \ \ \ (n)\]\[n=\frac{2mq+p}{2}\]
 Q16\[a=\frac{2b+c}{3c} \ \ \ \ \ (b)\]\[b=\frac{3acc}{2}\]
 Q17\[e=\frac{3(f^2g)}{2h} \ \ \ \ \ (f)\]\[f=\sqrt{\frac{2eh}{3}+g}\]
 Q18\[P=\sqrt{\frac{stuv}{w}} \ \ \ \ \ (w)\]\[w=\frac{stuv}{p^2}\]
 Q19\[\frac{1}{f}= \frac{1}{u}+\frac{1}{v} \ \ \ \ \ (f)\]\[f=\frac{uv}{u+v}\]
 Q20\[\frac{1}{f}= \frac{1}{u}+\frac{1}{v} \ \ \ \ \ (u)\]\[u=\frac{fv}{vf}\]
Exercise 3: Transposition of formulas #3 (Self inverse operations)
 Q1
Make $t$ the subject of the formula, $a=4t$.
$t=4a$  Q2
Make $x$ the subject of the formula, $p = t  x$.
$x=tp$  Q3
Express $d$ in terms of $a$ and $c$ in the formula, $c = a^2  d$.
$d=a^2c$  Q4
Express $p$ in terms of $q$ and $r$ in the formula, $q=\sqrt{r^2p}$.
$p=r^2q^2$  Q5
Make $m$ the subject of the formula, $p=\sqrt{n^2m^2}$.
$m=\pm \sqrt{n^2p^2}$  Q6
Make $a$ the subject of the formula, \[c=\frac{2}{a}\]
$a=\frac{2}{c}$  Q7
Make $q$ the subject of the formula, \[r=\frac{p}{q}\]
$q=\frac{p}{r}$  Q8
Express $t$ in terms of $u$ and $v$ using the formula, \[v=\frac{u^2}{t}\]
$t=\frac{u^2}{v}$  Q9
Express $m$ in terms of $n$, $p$ and $q$, using the formula, \[n=\frac{p+q}{m}\]
$m=\frac{p+q}{n}$  Q10
Make $x$ the subject of the formula, \[y=\frac{pq}{2+\sqrt{x}}\]
$x = {\left( {\frac{{p  q  2y}}{y}} \right)^2}$
Exercise 4: Transposition of formulas #4 (Really hard ones!)
 Q1
Make $m$ the subject of $I = mv  mu$.
\[m = \frac{I}{{v  u}}\]  Q2
Rearrange $y + x = 3(x + 2)$ to make $x$ the subject.
\[x = \frac{{y  6}}{2}\]  Q3
Make $x$ the subject of the formula $3(x + 2y) = 2(4x  3)$.
\[x = \frac{{6(1 + y)}}{5}\]  Q4
Make $a$ the subject of $c(a + b) = a  b$.
\[a = \frac{{b(1 + c)}}{{1  c}}\]  Q5
Make $q$ the subject of \[=\frac{q+1}{q1}\]
\[q = \frac{{r + 1}}{{r  1}}\]  Q6
Make $a$ the subject of \[2b=\frac{ac}{2a+c}\]
\[a = \frac{{c(1 + 2b)}}{{1  4b}}\]  Q7
Rearrange \[m=\frac{2(ns)}{3(n+s)}\] to make $n$ the subject.
\[n = \frac{{s(2 + 3m)}}{{2  3m}}\]  Q8
Rearrange \[m=\frac{2(ns)}{3(n+s)}\]to make $s$ the subject.
\[s = \frac{{n(2  3m)}}{{3m + 2}}\]  Q9
Make $p$ the subject of $r (pq + 7) = 2 (5p  2q)$.
\[p = \frac{{4q + 7r}}{{10  qr}}\]  Q10
Make $m$ the subject of \[E=\frac{1}{2}mv^2+mgh\]
\[m = \frac{{2E}}{{{v^2} + 2gh}}\]
Exercise 5: Word problems involving formulas
 Instructions:
Read the questions very carefully before attempting them. It is important to read the problem more than once, as you will miss important information, if you do not.
 Q1
The formula $v^2 = u^2 + 2as$ is used to find the velocity ($v$) when initial velocity ($u$), acceleration ($a$) and the displacement ($s$) are all known.
Change the subject of this formula, in each case, to the following letters:
 $v$
 $u$
 $a$
 \[v = \sqrt {{u^2} + 2as} \]
 \[u = \sqrt {{v^2}  2as} \]
 \[a = \frac{{{v^2}  {u^2}}}{{2s}}\]
 Q2
A recipe for roasting chicken is given as $45$ minutes per kg and then $20$ minutes extra.
 Using the letters $m$ for the mass of the chicken and $t$ for the time in minutes, make up a formula which could be used to find the cooking time (in minutes) for any roast chicken.
 Use your formula to find the cooking time in minutes for a chicken with a mass of 2.3kg.
 Change the subject of your formula to give a new formula which would allow you to find the mass of the chicken if you know the cooking time.
 Use your new formula to find the mass of a chicken which would have a cooking time of $1$ hour and $50$ minutes.
 \[t = 45m + 20\]
 $123.5$ minutes
 \[m = \frac{{t  20}}{{45}}\]
 $2$ kg
 Q3
A number $p$ is equal to the sum of the square of $q$ and the product of $3$ and $r$.
 Construct a formula for $p$ in terms of $q$ and $r$.
 Use your formula to find the value of $p$ when $q = 3$ and $r = 5$.
 Change the subject of the formula to make $q$ the subject.
 Use this formula to find the two possible values of $q$ when $p = 48$ and $r = 4$.
 Change the subject of the formula to make $r$ the subject.
 Use your formula to find the value of $r$ when $p = 76$ and $q = 5$.
 $p = q^2 + 3r$
 $24$
 \[q = \pm \sqrt {p  3r} \]
 $\pm 6$
 \[r = \frac{{p  {q^2}}}{3}\]
 $17$
 Q4
A number $a$ is defined as the sum of the cube of $b$ and the product of $5$ and $c^2$.
 Find a formula for $a$ in terms of $b$ and $c$.
 Use your formula to find the value of $a$ when $b = 2$ and $c = 3$.
 Change the subject of the formula to make $c$ the subject.
 Use your new formula to find the possible values of $c$ when $a = 133$ and $b = 2$.
 \[a = {b^3} + 5{c^2}\]
 $53$
 \[c = \pm \sqrt {\frac{{a  {b^3}}}{5}} \]
 $\pm 5$
 Q5
A number $x$ is equal to the square root of the sum of $p$ and twice $q$, all divided by $r$.
 Construct a formula for $x$ in terms of $p$, $q$ and $r$.
 Use your formula to find the value of $x$ when $p = 5$, $q = 2$ and $r = 2$.
 Rearrange the formula to make $p$ the subject.
 Use your new formula to find the value of $p$ when $x = 4$, $r = 3$ and $q = 1$.
 \[x = \frac{{\sqrt {p + 2q} }}{r}\]
 $1.5$
 \[p = {\left( {xr} \right)^2}  2q\]
 $142$
 Q6
A hire company rents out lawn mowers. The rental charge involves a flat fee of $£20$ and then $£12$ per day that the mower is hired out.
 Using $d$ to represent the number of days and $C$ to represent the total rental charge, form an equation connecting $C$ and $d$.
 Use your formula to find the amount charged to a customer who wishes to keep the mower for
 $1$ day
 $5$ days
 $1$ week
 $C=20+12d$

 $£32$
 $£80$
 $£104$
 Q7
A taxi service calculates their fares using a flat fee of $£3.50$ and then a charge of $£2.50$ per kilometre thereafter.
 Form an equation to calculate the fare ($£F$) in terms of the distance traveled ($d$ km), using this information.
 Use your formula to calculate the fare for journeys of
 $2$ km
 $18$ km
 $56$ km
 $F=3.5+2.5d$

 $£8.50$
 $£48.50$
 $£143.50$
 Q8
A furniture company calculates the cost of delivery using the following rules:
 Deliveries less than $5$ km are charged at $£2$ per km or part thereof.
 Deliveries between $5$ km and $10$ km inclusive, are charged a flat fee of $£12$ and then $£1.50$ per km or part thereof.
 Deliveries over $10$ km are charged a flat fee of $£15$ and then $£1.50$ per km or part thereof.
Use these rules to work out the cost of delivering goods to destinations which are these distances from the shop.
 $3$ km
 $6.8$ km
 $15.3$ km
 $£6.00$
 $£22.20$
 $£37.95$