Question
Derive the expression for average acceleration and instantaneous acceleration for the motion of an object in $x-y$ plane.
✓
Answer
Consider an object moving in an $x-y$ plane.
Let the velocity of the particle be $\vec{v}_1$ and $\vec{v}_2$ at time $t_1$ and $t_2$ respectively.
$ii)$ The average acceleration $\left(\overrightarrow{ a }_{ av }\right)$ of the particle between $t_1$ and $t_2$ is given as,
$\vec{a}_{ av }=\frac{\vec{v}_2-\vec{v}_1}{t_2-t_1}=\left(\frac{v_{2 x}-v_{1 x}}{t_2-t_1}\right) \hat{i}+\left(\frac{v_{2 y}-v_{1 y}}{t_2-t_1}\right) \hat{j}$
$\therefore \overrightarrow{a_{a v}}=\left(a_{a v}\right)_x \hat{i}+\left(a_{a v}\right)_y \hat{j}$
$iii.$ Thus, the $x$ and $y$ components of the average acceleration are given by,
$\left(a_{ av }\right)_{ x }=\left(\frac{ v _{2 x }- v _{1 x }}{ t _2- t _1}\right) \text { and }\left( a _{ av }\right)_y=\left(\frac{ v _{2 y }- v _{ 1y }}{ t _2-t_1}\right)$
$iv.$ The magnitude and direction of the average acceleration are given by,
$a _{ av }=\sqrt{\left( a _{ av }\right)_{ x }^2+\left( a _{ av }\right)_{ y }^2}$
$\tan \theta=\frac{\left( a _{ av }\right)_{ y }}{\left( a _{ xv }\right)_{ x }}$
$v.$ The instantaneous acceleration $(\vec{a})$ of the particle is given by, $\vec{a}=\lim _{t \rightarrow 0}\left(\frac{\Delta \vec{v}}{\Delta t}\right)=\frac{d \vec{v}}{d t}$
$\therefore \vec{a}=\left(\frac{d v_x}{d t}\right) \hat{i}+\left(\frac{d v_y}{d t}\right) \hat{j}$
$\therefore \vec{a}=\frac{d}{d t}\left(\frac{d x}{d t}\right) \hat{i}+\frac{d}{d t}\left(\frac{d y}{d t}\right) \hat{j}=\left(\frac{d^2 x}{d t^2}\right) \hat{i}+\left(\frac{d^2 y}{d t^2}\right) \hat{j}$
$vi.$ Thus, the $x$ and $y$ components of the instantaneous acceleration are given by,
$a_x=\frac{d^2 x}{d t^2}$ and $a_y=\frac{d^2 y}{d t^2}$
$vii.$ The magnitude and direction of instantaneous acceleration are given by,
$a=\sqrt{a_x^2+a_y^2}=\sqrt{\left(\frac{d^2 x}{d t^2}\right)^2+\left(\frac{d^2 y}{d t^2}\right)^2}$
$\tan \theta=\frac{a_y}{a_x}=\frac{\left(\frac{d v_y}{d t}\right)}{\left(\frac{d v_x}{d t}\right)}=\frac{d v_y}{d v_x}\ \ \ ....(1)$
Equation $(1)$ is the slope of the tangent to the curve at the point at which we are calculating the instantaneous acceleration.
Let the velocity of the particle be $\vec{v}_1$ and $\vec{v}_2$ at time $t_1$ and $t_2$ respectively.
$ii)$ The average acceleration $\left(\overrightarrow{ a }_{ av }\right)$ of the particle between $t_1$ and $t_2$ is given as,
$\vec{a}_{ av }=\frac{\vec{v}_2-\vec{v}_1}{t_2-t_1}=\left(\frac{v_{2 x}-v_{1 x}}{t_2-t_1}\right) \hat{i}+\left(\frac{v_{2 y}-v_{1 y}}{t_2-t_1}\right) \hat{j}$
$\therefore \overrightarrow{a_{a v}}=\left(a_{a v}\right)_x \hat{i}+\left(a_{a v}\right)_y \hat{j}$
$iii.$ Thus, the $x$ and $y$ components of the average acceleration are given by,
$\left(a_{ av }\right)_{ x }=\left(\frac{ v _{2 x }- v _{1 x }}{ t _2- t _1}\right) \text { and }\left( a _{ av }\right)_y=\left(\frac{ v _{2 y }- v _{ 1y }}{ t _2-t_1}\right)$
$iv.$ The magnitude and direction of the average acceleration are given by,
$a _{ av }=\sqrt{\left( a _{ av }\right)_{ x }^2+\left( a _{ av }\right)_{ y }^2}$
$\tan \theta=\frac{\left( a _{ av }\right)_{ y }}{\left( a _{ xv }\right)_{ x }}$
$v.$ The instantaneous acceleration $(\vec{a})$ of the particle is given by, $\vec{a}=\lim _{t \rightarrow 0}\left(\frac{\Delta \vec{v}}{\Delta t}\right)=\frac{d \vec{v}}{d t}$
$\therefore \vec{a}=\left(\frac{d v_x}{d t}\right) \hat{i}+\left(\frac{d v_y}{d t}\right) \hat{j}$
$\therefore \vec{a}=\frac{d}{d t}\left(\frac{d x}{d t}\right) \hat{i}+\frac{d}{d t}\left(\frac{d y}{d t}\right) \hat{j}=\left(\frac{d^2 x}{d t^2}\right) \hat{i}+\left(\frac{d^2 y}{d t^2}\right) \hat{j}$
$vi.$ Thus, the $x$ and $y$ components of the instantaneous acceleration are given by,
$a_x=\frac{d^2 x}{d t^2}$ and $a_y=\frac{d^2 y}{d t^2}$
$vii.$ The magnitude and direction of instantaneous acceleration are given by,
$a=\sqrt{a_x^2+a_y^2}=\sqrt{\left(\frac{d^2 x}{d t^2}\right)^2+\left(\frac{d^2 y}{d t^2}\right)^2}$
$\tan \theta=\frac{a_y}{a_x}=\frac{\left(\frac{d v_y}{d t}\right)}{\left(\frac{d v_x}{d t}\right)}=\frac{d v_y}{d v_x}\ \ \ ....(1)$
Equation $(1)$ is the slope of the tangent to the curve at the point at which we are calculating the instantaneous acceleration.
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