Explain average acceleration and instantaneous acceleration.

Average acceleration is defined as the time rate of change of velocity over a given time interval.
$\text{Average acceleration} = \frac{\text{Change in velocity}}{\text{Time interval}}$
The average acceleration $\vec{a}$ of an object moving in the $xy$-plane for a time interval $\Delta t$ is the change in velocity divided by the time interval:
$\vec{a} = \frac{\overrightarrow{\Delta v}}{\Delta t} = \frac{\Delta(v_x \hat{i} + v_y \hat{j})}{\Delta t} = \frac{\Delta v_x}{\Delta t} \hat{i} + \frac{\Delta v_y}{\Delta t} \hat{j} = a_x \hat{i} + a_y \hat{j}$
Instantaneous acceleration is the limiting value of the average acceleration as the time interval approaches zero:
$\vec{a} = \lim_{\Delta t \rightarrow 0} \frac{\overrightarrow{\Delta v}}{\Delta t} = \frac{d\vec{v}}{dt}$
Since $\vec{v} = v_x \hat{i} + v_y \hat{j}$,we have:
$\vec{a} = \frac{d}{dt}(v_x \hat{i} + v_y \hat{j}) = \frac{dv_x}{dt} \hat{i} + \frac{dv_y}{dt} \hat{j} = a_x \hat{i} + a_y \hat{j}$
where $a_x = \frac{dv_x}{dt}$ and $a_y = \frac{dv_y}{dt}$.
Furthermore,since $\vec{v} = \frac{d\vec{r}}{dt}$,acceleration can be expressed as the second derivative of position with respect to time:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right) = \frac{d^2\vec{r}}{dt^2} = \ddot{\vec{r}}$