These equations are valid only when the acceleration is constant. An example of constant acceleration motion is freely falling bodies under the action of gravity neglecting the air resistance.

If you are here to simply take a look at these equations, here are four kinematic equations derived here in original form (trust me you are going to use these equations a lot of times):

### Equations of straight line motion in their original form, also derived below

If you need to learn how these equations are derived, simply scroll down (you can also find a surprise!). In this summary, $$x_i$$ is initial position, $$x_f$$ is final position and therefore the displacement is $$x_f - x_i$$. Similarly, $$v_{xi}$$ is initial velocity, $$v_{xf}$$ is final velocity, $$a_x$$ is acceleration and $$t$$ is time period. The subscript $$x$$ simply denotes the motion is along x-axis of our coordinate system.

\begin{align*}{{v}_{xf}}&={{v}_{xi}}+a_xt \\{x_f} &= x_i + {{v}_{xi}}t+\frac{1}{2}{a_x}{{t}^{2}} \\\quad {{v}_{xf}}^{2}&={{v}_{xi}}^{2}+2{a_x}s \\{x_f}&=x_i+\left( \frac{{{v}_{f}}+{{v}_{i}}}{2} \right)t \end{align*}

### Equations of straight line motion in suvat form

The SUVAT equations are the kinematic equations for constant acceleration but in different notations of quantities involved. In this form, $$s$$ is displacement, $$u$$ is initial velocity, $$v$$ is final velocity, $$a$$ is acceleration and $$t$$ is time period. So if you combine letters $$s$$, $$u$$, $$v$$, $$a$$ and $$t$$ you get "SUVAT", therefore also known as SUVAT equations.

\begin{align*}v &= u+at \\ s &=ut+\frac{1}{2}{a}{{t}^{2}} \\\quad v^2&=u^{2}+2as \\s&=\left( \frac{v+u}{2} \right)t \end{align*}

## Let's derive

We first consider a situation of motion with constant acceleration. The figure below shows a car is moving in a straight line with constant acceleration. If you build a coordinate system for the motion of the car, let the motion is along x-axis (x-direction).

The acceleration is the rate of change of velocity. In acceleration $${a_x}$$ along x-axis. We know the equation for acceleration $${a_x}$$ if the final velocity is $${{v}_{xf}}$$ and initial velocity is $${{v}_{xi}}$$ as,

$a_x=\frac{{{v}_{xf}}-{{v}_{xi}}}{t} \tag{1} \label{1}$

The time $$t$$ in Equation \eqref{1} is the time period between the initial and final velocities. Note that the subscript $$x$$ means along x-axis and, $$i$$ and $$f$$ mean initial and final respectively. Simplifying Equation \eqref{1} gives,

${{v}_{xf}}={{v}_{xi}}+a_xt \tag{2} \label{2}$

The next step is to define average velocity:

${{v}_{x-\text{av}}}=\frac{{{v}_{xf}}+{{v}_{xi}}}{2} \tag{3} \label{3}$

In Figure 1 the magnitude of initial displacement of the car at point $${{p}_{1}}$$ is $${x_i}$$ and final displacement at point $${{p}_{2}}$$ is $${x_f}$$, the average velocity between these two points is, ${{v}_{x-\text{av}}}=\frac{{x_f}-{x_i}}{t} \tag{4} \label{4}$And again $$t$$ is the time period during the motion between the two points. Now it's time to equate Equation \eqref{3} with Equation \eqref{4} as both of these equations represent the same average velocity. $\frac{{{v}_{xf}}+{{v}_{xi}}}{2}=\frac{{x_f}-{x_i}}{t} \tag{5} \label{5}$ Putting the value of $${{v}_{xf}}$$ from Equation \eqref{2} in Equation \eqref{5} and simplifying we get, ${x_f}-{x_i}={{v}_{xi}}t+\frac{1}{2}{a_x}{{t}^{2}} \tag{6} \label{6}$ The quantity $${x_f}-{x_i}$$ is the displacement, Equation \eqref{6} can be rewritten as, $x_f=x_i+{{v}_{xi}}t+\frac{1}{2}{a_x}{{t}^{2}} \tag{7} \label{7}$Now replacing $$t$$ in Equation \eqref{7} by $$t=\frac{{{v}_{xf}}-{{v}_{xi}}}{a_x}$$ from Equation \eqref{1} and simplifying we get another important equation, \begin{align*} x_f - x_i&={{v}_{xi}}\left( \frac{{{v}_{xf}}-{{v}_{xi}}}{a_x} \right)+\frac{1}{2}a_x{{\left( \frac{{{v}_{xf}}-{{v}_{xi}}}{a_x} \right)}^{2}} \\ \text{or,}\quad {{v}_{xf}}^{2}&={{v}_{xi}}^{2}+2{a_x}(x_f - x_i) \tag{8} \label{8} \end{align*}Next we can find another useful equation by simplifying Equation \eqref{5} which is, ${x_f}-{x_i}=\left( \frac{{{v}_{f}}+{{v}_{i}}}{2} \right)t \tag{9} \label{9}$The equations of straight line motion with constant acceleration are Equation \eqref{2}, Equation \eqref{6}, Equation \eqref{8} and Equation \eqref{9}. These equations are used to solve problems related to straight line motion with constant acceleration. These equations are valid only when the acceleration is constant.

Sometimes what happens is, it seems reasonable but wrong. You might tempt to write $$v_{xf}=\frac{{x_f}-{x_i}}{t}$$ and put the this value in Equation \eqref{1} which becomes, ${x_f}-{x_i}={{v}_{i}}t+a{{t}^{2}}\text{ }$What's wrong here? You should know that $$v_{xf}$$ is the instantaneous velocity but the velocity given by $$\frac{{x_f}-{x_i}}{t}$$ is the average velocity. Notice the acceleration is constant and the velocity is not the same. So the average velocity and the instantaneous velocity are not the same. If you do this, you'll get a wrong equation.

## Let's derive the same equations using calculus

You can also derive the above kinematic equations with calculus if you are familiar with integration. Here it is assumed that you understand calculus.

In the first step of deriving with calculus we use the definition of acceleration, that is the rate of change of velocity is acceleration, so

$a_x = \frac{dv_x}{dt}$

You know that $$dv_x$$ is the infinitesimally small change in velocity. You can express $$v_{xf} - v_{fi}$$ as the integral, that is

$v_{xf} - v_{fi}= \int_{0}^{t} a_xdt$

Since the acceleration is constant, you can take $$a_x$$ out of the integral and finally get,

$v_{xf} = v_{fi} + a_x t \tag{10} \label{10}$

which is Equation \eqref{2} we derived earlier. Now we use the definition of velocity, that is the rate of change of displacement is velocity (note that we used the definition of acceleration to derive the above equation).

$v_x = \frac{dx}{dt}$

It is obvious that $$dx = v_xdt$$ and similarly you can turn the equation into integral form as

$x_f - x_i = \int_0^t v_xdt$

You already know the expression of velocity from Equation \eqref{10} and therefore replacing $$v_x$$ in above equation by $$v_f$$ gives

\begin{align*}x_f - x_i &= \int_0^t(v_{xi} + a_xt)dt \\\text{or,}\,\,\,\, x_f - x_i &= v_{xi} \int_0^tdt + a_x\int_0^t tdt \\\text{or,}\,\,\,\, x_f - x_i &= v_{xi}t + \frac{1}{2}a_xt^2 \tag{11} \label{11}\\\end{align*}

Note in the above integration that the initial velocity is constant and taken out of the integral. The Equation \eqref{11} just derived is the same as Equation \eqref{6} derived without calculus. Once you have these two major Equations \eqref{10} and \eqref{11} in place, the derivation of other equations is straightforward.