Transformation of formulae is necessary to calculate unknown values. Whatever is done to one side of a formula must be done to the other side. Examples show transforming formulas to make the unknown value the subject by adding or subtracting terms from both sides, or dividing both sides. Common formulas in mechanics like F=ma, SUVAT equations for distance, velocity, acceleration, and time are transformed to solve for unknowns by inserting known values.

1. Transformation of formulae
When given a formulae to work out an unknown you’ll notice allot of the time the value you are
tryingto workout isnot the subject (the value the equationequalsto) of the formula. This meaning
in order to find the unknown value, the equation must be re-arranged to make the value you are
trying to find the subject. The most important rule when it comes to transforming/re-arranging a
formula is to remember whatever is done to the one side must be done to the other.
Here is a very basic example of a question where you would need to transform the formula to
calculate the unknown value:
Question
Using the equation a + b = c, where a = 2 and c = 5 calculate the value of b.
Answer
 By looking at the formula you can see that c is the subject but because we need to find a
value forb we must make b the subject. We can do this by taking away a from both sides to
give us:
b = c – a
 Nowwe have transformedthe formulagiventomake bthe subject we can insert our values
for a and c to give us a value for b.
b = 5 – 2
b = 3
A formula that is commonly used in mechanics and may be needed in an exam is F = ma, where
F = force, m = mass and a = acceleration.
Here is a question where F = ma is needed:
Question
What is the mass of a falling rock if it produces a force of 147 N?
Answer
 Using the information given we know the rock is falling therefore the acceleration of the
rock is g which is 9.8ms-2
(2s.f.) and it tells us the force produced by the rock is 147N,
therefore we
have F and a.
 Nextwe needtotransformthe formula F= ma to make m the subject. We do thisby dividing
both sides by a.
F = ma
m = F/a

2.  Finally by inserting the known values into the formula we can calculate the mass of the
falling rock.
m = (147)/(9.8)
m = 1440.6kg
Another set of formulae which are used in mechanics and will more than likely be needed in an
exam are the SUVAT formulae:
Where:
 S = Distance (m)
 U = Initial velocity (ms-1
)
 V = Final velocity (ms-1
)
 A = Acceleration (ms-2
)
 T = Time (s)
Here are some examples of a SUVAT questions where transforming the formulae is needed:
Question
A particle is accelerated uniformly from rest, so that after 10 seconds it has achieved a speed of 15
ms-1
, find its acceleration.
Answer
 The first step to answering a question like this is to record what values are given.
u = 0ms-1
v = 15ms-1
t = 10s
 Next we look at what equation we can use to calculate the acceleration using the values
given.
v = u + at.

3.  Now we re-arrange the formula to make a the subject.
v = u + at
v-u = at
a = (v-u)/t
 Lastly we insert the known values into the formula to get a value for a.
a = (15-0)/10
a = 1.5ms-2
Here is another SUVAT question:
Question
A train starts from rest and accelerates uniformly at 1.5ms-2
until it attains a speed of 30ms-1
. Find
the distance travelled.
Answer
 Like the first SUVAT question we need to make a note of the known values.
u = 0ms-1
v = 30ms-1
a = 1.5ms-1
 Next we select the equation needed to find the distance travelled using the know values.
V2
= u2
+ 2as
 Now we need to transform the equation to make s the subject.
v2
= u2
+ 2as
v2
- u2
= 2as
s = (v2
- u2
)/2a
 Finally we input our known values to get a value for s.
s = (302
– 02)/2(1.5)
s = 300m