In this article you'll learn how to derive equations of straight line motion based on constant acceleration. It means these equations are valid only when the acceleration is constant.

The acceleration is the rate of change of velocity. In Figure 1 a car is moving with constant acceleration ${a_x}$ along x-axis. Here we are discussing the straight line motion with constant acceleration and we know the equation for acceleration ${a_x}$ if the final velocity is ${{v}_{fx}}$ and initial velocity is ${{v}_{ix}}$ as, \[a=\frac{{{v}_{fx}}-{{v}_{ix}}}{t} \tag{1} \label{1}\]

The time $t$ in Eq. \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 Eq. \eqref{1} gives,

\[{{v}_{fx}}={{v}_{ix}}+at \tag{2} \label{2}\]

The next step is to define average velocity:

\[{{v}_{av}}=\frac{{{v}_{fx}}+{{v}_{ix}}}{2} \tag{3} \label{3}\]

In Figure 1 the magnitude of initial displacement of the car at point ${{p}_{1}}$ is ${{s}_{i}}$ and final displacement at point ${{p}_{2}}$ is ${{s}_{f}}$, the average velocity between these two points is,

\[{{v}_{av}}=\frac{{{s}_{f}}-{{s}_{i}}}{t} \tag{4} \label{4}\]

And again $t$ is the time period during the motion between the two points. Now it's time to equate Eq. \eqref{3} with Eq. \eqref{4} as both of these equations represent the same average velocity.

\[\frac{{{v}_{fx}}+{{v}_{ix}}}{2}=\frac{{{s}_{f}}-{{s}_{i}}}{t} \tag{5} \label{5}\]

Putting the value of ${{v}_{fx}}$ from Eq. \eqref{2} in Eq. \eqref{5} and simplifying we get,

\[{{s}_{f}}-{{s}_{i}}={{v}_{ix}}t+\frac{1}{2}{a_x}{{t}^{2}} \tag{6} \label{6}\]

If $s={{s}_{f}}-{{s}_{i}}$, Eq. \eqref{6} becomes,

\[s={{v}_{ix}}t+\frac{1}{2}{a_x}{{t}^{2}} \tag{7} \label{7}\]

Now replacing $t$ in Eq. \eqref{7} by $t=\frac{{{v}_{fx}}-{{v}_{ix}}}{a}$ from Eq. \eqref{1} and simplifying we get another important equation,

\[\begin{align*} s&={{v}_{ix}}\left( \frac{{{v}_{fx}}-{{v}_{ix}}}{a_x} \right)+\frac{1}{2}a{{\left( \frac{{{v}_{fx}}-{{v}_{ix}}}{a_x} \right)}^{2}} \\ \text{or,}\quad {{v}_{fx}}^{2}&={{v}_{ix}}^{2}+2{a_x}s \tag{8} \label{8} \end{align*}\]

Next we can find another useful equation by simplifying Eq. \eqref{5} which is,

\[{{s}_{f}}-{{s}_{i}}=s=\left( \frac{{{v}_{f}}+{{v}_{i}}}{2} \right)t \tag{9} \label{9}\]

The equations of straight line motion with constant acceleration are Eq. \eqref{2}, Eq. \eqref{6}, Eq. \eqref{8} and Eq. \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 ${{v}_{f}}=\frac{{{s}_{f}}-{{s}_{i}}}{t}$ and put the value of ${{v}_{f}}=\frac{{{s}_{f}}-{{s}_{i}}}{t}$ in Eq. \eqref{1} which becomes,

\[{{s}_{f}}-{{s}_{i}}={{v}_{i}}t+a{{t}^{2}}\text{ }\]

What's wrong here? You should know that ${{v}_{f}}$ is the instantaneous velocity but the velocity given by $\frac{{{s}_{f}}-{{s}_{i}}}{t}$ 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.