
For constant acceleration, you have now learned two equations of motion: equations (4.2) and (4.3).
These two equations can be used to solve a wide variety of motion problems in physics.
In some problems, however, it might be more convenient to use two additional equations.
These two additional equations aren’t unique or different from the other ones, they are just more convenient to use in certain circumstances. As you will see below, we get the two new equations by manipulating the two equations you already know!

Start with equation (4.3):
$$x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}$$ 
Assuming that we are starting at an initial time t = 0, acceleration is the change in velocity divided by the change in time or a = (v − v_{0})/t. Use this equation to substitute for acceleration in equation (4.3) to get
$$\begin{array}{lll}x\hfill & =\hfill & {x}_{0}+{v}_{0}t+\frac{1}{2}\left(\frac{v{v}_{0}}{t}\right){t}^{2}\hfill \\ \hfill & =\hfill & {x}_{0}+{v}_{0}t+{\scriptscriptstyle \frac{1}{2}}vt{\scriptscriptstyle \frac{1}{2}}{v}_{0}t\hfill \\ \hfill & =\hfill & {x}_{0}+{\scriptscriptstyle \frac{1}{2}}\left({v}_{0}+v\right)t\hfill \end{array}$$

This third equation of motion is useful when you don’t know the acceleration.

(4.4)  $$x={x}_{0}+{\scriptscriptstyle \frac{1}{2}}\left({v}_{0}+v\right)t$$
 x  =  position (m)  x_{0}  =  initial position (m/s)  v_{0}  =  initial velocity (m/s)  v  =  velocity (m/s)  t  =  time (s) 
 Position when acceleration
is not known 

The third equation can also be expressed as
$$x{x}_{0}={\scriptscriptstyle \frac{1}{2}}\left({v}_{0}+v\right)t$$
The lefthand side of this equation is the displacement, or the final position minus the initial position.
On the righthand side of this equation is the quantity ½(v_{0} + v), which is the average velocity when the acceleration is constant.
So the righthand side is the average velocity multiplied by the time. Equation (4.4) can therefore be described as “displacement equals the average velocity multiplied by the time interval.”

For the fourth equation of motion, start with equation (4.4) above, but rewrite it as
$$\begin{array}{ccc}x{x}_{0}={\scriptscriptstyle \frac{1}{2}}\left({v}_{0}+v\right)t& \Rightarrow & 2\left(x{x}_{0}\right)=\left({v}_{0}+v\right)t\end{array}$$

This time we want to remove the dependence on time. Rewrite the definition of acceleration to solve for time
$$\begin{array}{ccc}a=\frac{v{v}_{0}}{t}& \Rightarrow & t=\frac{v{v}_{0}}{a}\end{array}$$

and substitute this equation for acceleration to get
$$\begin{array}{ccccc}2\left(x{x}_{0}\right)=\left({v}_{0}+v\right)\left(\frac{v{v}_{0}}{a}\right)& \Rightarrow & 2\left(x{x}_{0}\right)=\frac{{v}^{2}{v}_{0}^{2}}{a}& \Rightarrow & 2a\left(x{x}_{0}\right)={v}^{2}{v}_{0}^{2}\end{array}$$

Rearranging these terms leads to equation (4.5), which is useful for solving problems when you don’t know the time.

(4.5)  $${v}^{2}={v}_{0}^{2}+2a\left(x{x}_{0}\right)$$
 v  =  velocity (m/s)  v_{0}  =  initial velocity (m/s)  a  =  acceleration (m/s^{2})  x  =  position (m)  x_{0}  =  initial position (m) 
 Velocity when time is
not known 

The fourth equation is often rewritten to put the velocity terms on one side of the equation:
$${v}^{2}{v}_{0}^{2}=2a\left(x{x}_{0}\right)$$

In this fourth equation, the change in the square of the velocity is related to the acceleration and the displacement (x − x_{0}).
As you will learn in Chapters 9 and 10, if you multiplied the equation by ½m, this expression relates changes in kinetic energy to the change in gravitational potential energy. The fourth equation is basically a statement of energy conservation when only kinetic and gravitational potential energy are considered.
Note that equation four does not include the variable for time. As you will discover in those later chapters, you can use energy conservation to solve physics problems without having to know or solve for time!

Equations (4.2), (4.3), (4.4), and (4.5) are the four equations of motion.
