Derive equation of motion of body moving in two dimensions overrightarrow v , = ,overrightarrow {{v

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3-2.Motion in Plane

hard

Derive equation of motion of body moving in two dimensions

$\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.

Option A

Option B

Option C

Option D

Solution

Consider a particle moving along $XY-$ direction with constant acceleration $\vec{a}$ and at time $t=0$, the velocity is initial velocity which is $\overrightarrow{v_{0}}$ and at time $t=t$, velocity is $\vec{v}$.

Since the particle is moving with constant acceleration, the instantaneous acceleration can be written as,

$\therefore \vec{a}=\frac{\vec{v}-\vec{v}_{0}}{t-0}$

$\therefore \vec{a}=\frac{\vec{v}-\vec{v}_{0}}{t}$

$\therefore \vec{v}=\overrightarrow{v_{0}}+\overrightarrow{a t}$

Writing this equation in terms of its components,

$v_{x}=v_{0 x}+a_{x} t$

$v_{y}=v_{0 y}+a_{y} t$

Let at $t=0$ the position vector of the particle is

given by $\overrightarrow{r_{0}}$ and at $t=t$ it is $\vec{r}$.

Standard 11

Physics

The velocity of a body at time $ t = 0$ is $10\sqrt 2 \,m/s$ in the north-east direction and it is moving with an acceleration of $ 2 \,m/s^{2}$ directed towards the south. The magnitude and direction of the velocity of the body after $5\, sec$ will be

hard

The position of a particle moving in the $xy-$plane at any time $t$ is given by $x = (3{t^2} – 6t)$ metres, $y = ({t^2} – 2t)$ metres. Select the correct statement about the moving particle from the following

medium

The position of a particle moving in the $xy-$ plane at any time $t$ is given by $x = (3t^2 -6t)\, metres$, $y = (t^2 -2t)\,metres$. Select the correct statement about the moving particle from the following

medium

A particle moves such that its position vector $\overrightarrow{\mathrm{r}}(\mathrm{t})=\cos \omega \mathrm{t} \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\overrightarrow{\mathrm{v}}(\mathrm{t})$ and acceleration $\overrightarrow{\mathrm{a}}(\mathrm{t})$ of the particle

medium

(JEE MAIN-2020)

The length of second's hand in watch is $1 \,cm.$ The change in velocity of its tip in $15$ seconds is

medium

Derive equation of motion of body moving in two dimensions $\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.

Solution

Similar Questions

The velocity of a body at time $ t = 0$ is $10\sqrt 2 \,m/s$ in the north-east direction and it is moving with an acceleration of $ 2 \,m/s^{2}$ directed towards the south. The magnitude and direction of the velocity of the body after $5\, sec$ will be

The position of a particle moving in the $xy-$plane at any time $t$ is given by $x = (3{t^2} – 6t)$ metres, $y = ({t^2} – 2t)$ metres. Select the correct statement about the moving particle from the following

The position of a particle moving in the $xy-$ plane at any time $t$ is given by $x = (3t^2 -6t)\, metres$, $y = (t^2 -2t)\,metres$. Select the correct statement about the moving particle from the following

The length of second's hand in watch is $1 \,cm.$ The change in velocity of its tip in $15$ seconds is

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Derive equation of motion of body moving in two dimensions

$\overrightarrow v \, = \,\overrightarrow {{v_0}} \, + \overrightarrow a t$ and $\overrightarrow r \, = \,\overrightarrow {{r_0}} \, + \overrightarrow {{v_0}} t\, + \,\frac{1}{2}g{t^2}$.