A particle projected from the origin moves in | vedclass

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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

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$y = x^2$

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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,$y = x^2$.

$A$ 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.

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