(A) Given velocity vector: $\vec{v} = v_x \hat{i} + v_y \hat{j} = 3 \hat{i} + 6x \hat{j}$.Comparing components,we get $v_x = \frac{dx}{dt} = 3$ and $v

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(A) Given velocity vector: $\vec{v} = v_x \hat{i} + v_y \hat{j} = 3 \hat{i} + 6x \hat{j}$.Comparing components,we get $v_x = \frac{dx}{dt} = 3$ and $v_y = \frac{dy}{dt} = 6x$.The slope of the path is given by $\frac{dy}{dx} = \frac{v_y}{v_x}$.Substituting the values,$\frac{dy}{dx} = \frac{6x}{3} = 2x$.Integrating both sides with respect to $x$: $\int dy = \int 2x dx$.Since the particle starts from the origin $(0,0)$,the constant of integration is $0$.Thus,$y = x^2$.

Class 11 Physics 3-2.Motion in Plane Motion In Two And Three Dimension

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$A$ particle projected from the origin moves in the $x-y$ plane with a velocity $\vec{v} = 3 \hat{i} + 6x \hat{j}$,where $\hat{i}$ and $\hat{j}$ are the unit vectors along the $x$ and $y$ axes. Find the equation of the path followed by the particle.

A
$y = x^2$
B
$y = \frac{1}{x^2}$
C
$y = 2x^2$
D
$y = \frac{1}{x}$

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3-2.Motion in Plane

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Motion In Two And Three Dimension

$A$ body moves $6\,m$ north,$8\,m$ east,and $10\,m$ vertically upwards. What is its resultant displacement from the initial position?

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The coordinates of a particle moving in the $XY$-plane vary with time as $x = 4t^2$ and $y = 2t$. The locus of the particle is a:

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$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

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The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 5t - 2t^2$ and $y = 10t$ respectively,where $x$ and $y$ are in meters and $t$ is in seconds. The acceleration of the particle at $t = 2 \, s$ is . . . . . . $m/s^2$.

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$A$ body starts from rest from the origin with an acceleration of $6 \; m/s^2$ along the $x$-axis and $8 \; m/s^2$ along the $y$-axis. Its distance from the origin after $4 \; s$ will be (in $; m$)

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$A$ particle projected from the origin moves in the $x-y$ plane with a velocity $\vec{v} = 3 \hat{i} + 6x \hat{j}$,where $\hat{i}$ and $\hat{j}$ are the unit vectors along the $x$ and $y$ axes. Find the equation of the path followed by the particle.

Similar Questions

$A$ body moves $6\,m$ north,$8\,m$ east,and $10\,m$ vertically upwards. What is its resultant displacement from the initial position?

The coordinates of a particle moving in the $XY$-plane vary with time as $x = 4t^2$ and $y = 2t$. The locus of the particle is a:

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 5t - 2t^2$ and $y = 10t$ respectively,where $x$ and $y$ are in meters and $t$ is in seconds. The acceleration of the particle at $t = 2 \, s$ is . . . . . . $m/s^2$.

$A$ body starts from rest from the origin with an acceleration of $6 \; m/s^2$ along the $x$-axis and $8 \; m/s^2$ along the $y$-axis. Its distance from the origin after $4 \; s$ will be (in $; m$)

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