“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-time graph for a car is semicircle as shown here. Which of the following is correct :
A particle starts from the origin at $\mathrm{t}=0$ with an initial velocity of $3.0 \hat{\mathrm{i}} \;\mathrm{m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \;\mathrm{m} / \mathrm{s}^{2} .$ The $\mathrm{x}$ -coordinate of the particle at the instant when its $y-$coordinate is $32\;\mathrm{m}$ is $D$ meters. The value of $D$ is
A particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}.$ Where $a, b$ and $c$ are contants. The acceleration of the particle is
$Assertion$ : If a body is thrown upwards, the distance covered by it in the last second of upward motion is about $5\, m$ irrespective of its initial speed
$Reason$ : The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped.