## 5. Transforming equations

Now that you understand the basic operations, try some more complex transformations that allow you to express one variable in terms of others. Starting with a simple equation linking speed (u), distance (s) and time (t), it is easy to see that, by using the rules you have already established:

if u = s/t

then s = ut

and t = s/u

This is for an object traveling at a constant speed. If the object is accelerating at a constant rate, its speed increases with time. If the initial speed is still denoted by u and its speed at a time t is represented by v, then the constant acceleration (a) is given by:

a = (v - u)/t

If you know the acceleration, you can transform this equation to find the value of v at a particular time, t:

a = (v - u)/t

at = v - u

v = u + at

From this relationship, you can find the distance traveled (s) in a given time (t). At the start of this section, you saw that distance is speed multiplied by time. When an object is under constant acceleration, the average speed at a time t is half of the sum of the initial speed (u) and the speed (v) at time t. So you can calculate distance traveled as:

s = 0.5(u + v)t

If you don't know the value of v but do know the value of a, you can replace the value of v in the last equation:

v = u + at

s = 0.5(u + v)t

s = 0.5(u + u + at)t

s = ut + 0.5 at^{2}

(Obviously, if the object is accelerating from rest, u is zero, and the equation becomes: s = 0.5 at^{2})

Alternatively, you can rearrange the same equation to find the value for v when you know u, a and s but not t. Time (t) can be expressed in terms of velocity and acceleration:

v = u + at

at = u - v

t = (v - u)/a

You can substitute this expression into the earlier equation:

s = ut + 0.5 at^{2}

s = u (v -u)/a + 0.5 a [(v - u)/a]^{2}

This looks quite daunting. It helps to work out the most complex expression separately. This is represented by a ‘stand-in’ variable, shown here as A:

s = u (v - u)/a + A

A = 0.5 a [(v - u)/a]^{2}

A = 0.5 a (v - u)2/a^{2}

The expression (v-u)^{2} means (v-u) multiplied by (v-u) so that A now becomes:

A = 0.5 a (v^{2} - 2uv + u^{2})/a^{2}

A = (0.5v^{2}/a) - (uv/a) + (0.5u^{2}/a)

Having dealt with the complex expression, it can be plugged back into the equation:

s = u (v - u)/a + A

s = u (v -u)/a + 0.5v^{2}/a - uv/a + 0.5u^{2}/a

s = uv/a - u^{2}/a + 0.5v^{2}/a - uv/a + 0.5u^{2}/a

as = uv - u^{2} + 0.5v^{2} - uv + 0.5u^{2}

as = 0.5v^{2} - 0.5u^{2}

v^{2} = u^{2} + 2as

This section has shown how the simple transforms presented in this document can be used to rearrange equations to derive new relationships between variables, to meet different needs. In this example, the main equations of motion under constant acceleration have been derived from simple expressions relating speed, distance and time. If you understood the much simpler equations earlier in this document, you should have been able to follow the different transformations in this section.

## Simultaneous equations

If you want to find out the value of an unknown, you need to know the value of everything else in the equation. So if you have an equation with two unknowns, like:

4x + 12y = 44

it is impossible to find unique solutions for x and y. However, if you have a second equation containing the same unknowns, you can use the same techniques as you have been using on single equations to find the values of both unknowns.

4x + 12y = 44

7x + 4y = 26

The technique used to find the values of the unknowns in these equations is quite simple. You need to change one or both of them so that the multiplier for one of the unknowns is the same in both equations. In this case it is easy to see that the multiplier for y in the second equation is exactly one-third of the value in the first. You know that you can keep the equation balanced if the same operation is performed on both sides. Here, we will multiply both sides by three:

3 × (7x + 4y) = 3 × 26

21x + 12y = 78

If we subtract one equation from the other, the y terms will cancel out, leaving only x terms. This will leave a simple equation in x.

21x + 12y = 78

4x + 12y = 44

subtraction leaves:

17x = 34

x = 2

The solution to x can then be added to either of the original equations to find y:

7x + 4y = 26

14 + 4y = 26

y = 3