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.

Figure 5 The velocity changes from ${{v}_1}$ to ${{v}_2}$. Only the magnitude of the velocity changes but the direction remains the same.

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.

Figure 6 The acceleration is constant and the velocity-time graph is a straight line.
Figure 7 The acceleration is not constant, therefore the velocity-time graph is not a straignt line. The slope gives the instantaneous acceleration at any point on the curve.

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

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