Transposing formulae 85
9. I = PRT
(T )
10. X L = 2π fL
(L)
11. I =
E
R
(R)
12. y =
x
a
+ 3
(x)
13. F =
9
5
C + 32
(C)
14. X C =
1
2π f C
( f )
12.3 Further transposing of formulae
Here are some more transposition examples to help
us further understand how more difficult formulae are
transposed.
Problem 9. Transpose the formula v = u +
Ft
m
to
make F the subject
v = u +
Ft
m
relates final velocity v, initial velocity u,
force F, mass m and time t .
F
m
is acceleration a.
Rearranging gives
u +
Ft
m
= v
and
Ft
m
= v − u
Multiplying each side by m gives
m
Ft
m
= m(v − u)
Cancelling gives
Ft = m(v − u)
Dividing both sides by t gives
Ft
t
=
m (v − u)
t
Cancelling gives F =
m (v − u)
t
or F =
m
t
(v − u)
This shows two ways of expressing the answer. There
is often more than one way of expressing a transposed answer. In this case, these equations for F are
equivalent; neither one is more correct than the other.
Problem 10. The final length L 2 of a piece of
wire heated through θ ◦ C is given by the formula
L 2 = L 1 (1 + αθ) where L 1 is the original length.
Make the coefficient of expansion α the subject
Rearranging gives
L 1 (1 + αθ) = L 2
Removing the bracket gives L 1 + L 1 αθ = L 2
Rearranging gives
L 1 αθ = L 2 − L 1
Dividing both sides by L 1 θ gives
L 1 αθ
L 1 θ
=
L 2 − L 1
L 1 θ
Cancelling gives
α =
L 2 − L 1
L 1 θ
An alternative method of transposing L 2 = L 1 (1 + αθ)
for α is:
Dividing both sides by L 1 gives
L 2
L 1
= 1 + αθ
Subtracting 1 from both sides gives
L 2
L 1
− 1 = αθ
or
αθ =
L 2
L 1
− 1
Dividing both sides by θ gives
α =
L 2
L 1
− 1
θ
The two answers α =
L 2 − L 1
L 1 θ
and α =
L 2
L 1
− 1
θ
look
quite different. They are, however, equivalent. The first
answer looks tidier but is no more correct than the
second answer.
Problem 11. A formula for the distance s moved
by a body is given by s =
1
2
(v + u)t . Rearrange the
formula to make u the subject
Rearranging gives
1
2
(v + u)t = s
Multiplying both sides by 2 gives
(v + u)t = 2s
Dividing both sides by t gives
(v + u)t
t
=
2s
t
Cancelling gives
v + u =
2s
t
Rearranging gives
u =
2s
t
− v or u =
2s − vt
t
Problem 12. A formula for kinetic energy is
k =
1
2
mv 2 . Transpose the formula to make v the
subject
Rearranging gives
1
2
mv
2
= k
9. I = PRT
(T )
10. X L = 2π fL
(L)
11. I =
E
R
(R)
12. y =
x
a
+ 3
(x)
13. F =
9
5
C + 32
(C)
14. X C =
1
2π f C
( f )
12.3 Further transposing of formulae
Here are some more transposition examples to help
us further understand how more difficult formulae are
transposed.
Problem 9. Transpose the formula v = u +
Ft
m
to
make F the subject
v = u +
Ft
m
relates final velocity v, initial velocity u,
force F, mass m and time t .
F
m
is acceleration a.
Rearranging gives
u +
Ft
m
= v
and
Ft
m
= v − u
Multiplying each side by m gives
m
Ft
m
= m(v − u)
Cancelling gives
Ft = m(v − u)
Dividing both sides by t gives
Ft
t
=
m (v − u)
t
Cancelling gives F =
m (v − u)
t
or F =
m
t
(v − u)
This shows two ways of expressing the answer. There
is often more than one way of expressing a transposed answer. In this case, these equations for F are
equivalent; neither one is more correct than the other.
Problem 10. The final length L 2 of a piece of
wire heated through θ ◦ C is given by the formula
L 2 = L 1 (1 + αθ) where L 1 is the original length.
Make the coefficient of expansion α the subject
Rearranging gives
L 1 (1 + αθ) = L 2
Removing the bracket gives L 1 + L 1 αθ = L 2
Rearranging gives
L 1 αθ = L 2 − L 1
Dividing both sides by L 1 θ gives
L 1 αθ
L 1 θ
=
L 2 − L 1
L 1 θ
Cancelling gives
α =
L 2 − L 1
L 1 θ
An alternative method of transposing L 2 = L 1 (1 + αθ)
for α is:
Dividing both sides by L 1 gives
L 2
L 1
= 1 + αθ
Subtracting 1 from both sides gives
L 2
L 1
− 1 = αθ
or
αθ =
L 2
L 1
− 1
Dividing both sides by θ gives
α =
L 2
L 1
− 1
θ
The two answers α =
L 2 − L 1
L 1 θ
and α =
L 2
L 1
− 1
θ
look
quite different. They are, however, equivalent. The first
answer looks tidier but is no more correct than the
second answer.
Problem 11. A formula for the distance s moved
by a body is given by s =
1
2
(v + u)t . Rearrange the
formula to make u the subject
Rearranging gives
1
2
(v + u)t = s
Multiplying both sides by 2 gives
(v + u)t = 2s
Dividing both sides by t gives
(v + u)t
t
=
2s
t
Cancelling gives
v + u =
2s
t
Rearranging gives
u =
2s
t
− v or u =
2s − vt
t
Problem 12. A formula for kinetic energy is
k =
1
2
mv 2 . Transpose the formula to make v the
subject
Rearranging gives
1
2
mv
2
= k