What if the velocity changes with time? Acceleration is the rate of change of velocity also called *linear* acceleration similar to the rate of change of displacement is velocity.

In Figure 5 a car has velocity ${{v}_{1}}$ at point ${{p}_{1}}$ and ${{v}_{2}}$ at point ${{p}_{2}}$ along x-axis. The earlier velocity ${{v}_{1}}$ is the initial velocity ${{v}_{i}}$ and the final one is ${{v}_{2}}$ which is ${{v}_{f}}$. The subscripts $i$ and $f$ represent the initial and final velocities respectively.

The change in velocity is the final velocity minus the initial velocity, that is, $\Delta v={{v}_{2}}-{{v}_{1}}={{v}_{f}}-{{v}_{i}}$. The time of the car at point $p_1$ is $t_1$ and at point $p_2$ is $t_2$. The time taken by the car to move from point $p_1$ to the point $p_2$ is $\Delta t={{t}_{2}}-{{t}_{1}}$. We express the rate of change in velocity as,

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

Notice that the subscript "$av$" represents the *average acceleration*. The acceleration in Eq. \eqref{4} is average acceleration because we don't know the rate of change of velocity is constant or not throughout the motion.

The whole process going here is, suppose the velocity at point ${{p}_{1}}$ is $5\text{m/s}$ and after $1\text{s}$ it is $10\text{m/s}$, and again after 1s it is $12\text{m/s}$. The acceleration during the first $1\text{s}$ is $\frac{(10-5)\text{m/s}}{1\text{s}}=5\text{m/}{{\text{s}}^{\text{2}}}$ and during the next $1\text{s}$ is $\frac{(12-10)\text{m/s}}{1\text{s}}=2\text{m/}{{\text{s}}^{\text{2}}}$, so the average of these two accelerations is $\frac{(5+2)\text{m/s}}{2}=3.5\text{m/}{{\text{s}}^{\text{2}}}$ which is the same as $\frac{(12-5)\text{m/s}}{2\text{s}}=3.5\text{m/}{{\text{s}}^{\text{2}}}$.

When $\Delta t$ approaches zero, the acceleration is *instantaneous acceleration* similar to instantaneous velocity. The instantaneous acceleration is the acceleration at a particular instant of time.

The instantaneous acceleration gives the acceleration at any point at any instant of time, so we mean the term instantaneous acceleration as a single word "acceleration". For example, if we say that a car has acceleration $3\text{m/}{{\text{s}}^{\text{2}}}$ at any point, that acceleration is instantaneous acceleration at that point.

The instantaneous acceleration is the average acceleration in Eq. \eqref{4} in the limit that $\Delta t$ approaches zero (the point ${{p}_{1}}$ is very close to the point ${{p}_{2}}$).

\[{{a}_{\text{ins}}}=a=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{{{v}_{f}}-{{v}_{i}}}{\Delta t}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta v}{\Delta t}=\frac{dv}{dt} \tag{5} \label{5}\]

The subscript $\text{ins}$ represents instantaneous for instantaneous acceleration which is not always necessary.

The graph of ${{v}_{x}}-t$ is always a straight line if the acceleration is constant indicating the constant slope. You may know that the slope at any point on the ${{v}_{x}}-t$ graph gives the instantaneous acceleration at that point. In Figure 6 the slope is constant and acceleration is the same (constant) throughout the motion.

What if the acceleration is not constant? You will find the result of ${{v}_{x}}-t$ graph to be some kind of curve and the slope at any point on the curve gives the acceleration at that point as in Figure 7