“Explain average acceleration and instantaneous acceleration.”
"The time rate of change of velocity for corresponding interval of time is called average acceleration."
$\text { Average acceleration }=\frac{\text { Change in velocity }}{\text { Time interval }}$
The average acceleration $\vec{a}$ of an object for a time interval $\Delta t$ moving in $x y$-plane is the change
in velocity divided by the time interval :
$\vec{a}=\frac{\overrightarrow{\Delta v}}{\Delta t}=\frac{\Delta\left(v_{x} \hat{i}+v_{y} \hat{j}\right)}{\Delta t}=\frac{\Delta v_{x}}{\Delta t} \hat{i}+\frac{\Delta v_{y}}{\Delta t} \hat{j} \quad \vec{a}=a_{x} \hat{i}+a_{y} \hat{j}ac$
The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero.
$\left(\vec{a}=\lim _{\Delta t \rightarrow 0} \frac{\overrightarrow{\Delta v}}{\Delta t}=\frac{\overrightarrow{d v}}{d t}\right)$
Since $\overrightarrow{\Delta v}=\Delta v_{x} \hat{i}+\Delta v_{y} \hat{j}$, we have
$\vec{a}=\hat{i} \lim _{\Delta t \rightarrow 0} \frac{\Delta v_{x}}{\Delta t}+\hat{j} \lim _{\Delta t \rightarrow 0} \frac{\Delta v_{y}}{\Delta t}=\frac{d v_{x}}{d t} \hat{i}+\frac{d v_{y}}{d t} \hat{j}$
$\vec{a}=a_{x} \hat{i}+a_{y} \hat{j}$
where, $a_{x}=\frac{d v_{x}}{d t}, a_{y}=\frac{d v_{y}}{d t}$
Equation $(1)$ represent that differentiation of velocity w.r.t. time gives acceleration.
$\vec{a}=\frac{\overrightarrow{d v}}{d t}=\frac{d}{d t}\left(\frac{\overrightarrow{d r}}{d t}\right)=\frac{d^{2} \vec{r}}{d t^{2}}=\ddot{\vec{r}}$
Equation$ (3)$ represent that double differentiation of position (displacement) w.r.t. time gives acceleration.
Velocity of a particle moving in a curvilinear path in a horizontal $X$ $Y$ plane varies with time as $\vec v = (2t\hat i + t^2 \hat j) \ \ m/s.$ Here, $t$ is in second. At $t = 1\ s$
The position vector of an object at any time $t$ is given by $3 t^2 \hat{i}+6 t \hat{j}+\hat{k}$. Its velocity along $y$-axis has the magnitude
Two particles $A$ and $B$ are moving in $XY$ plane. Particle $A$ moves along a line with equation $y = x$ while $B$ moves along $X$ axis such that their $X$ coordinates are always equal. If $B$ moves with a uniform speed $3\ m/s$ , the speed of $A$ is