A particle moves in the x-y plane with veloci | vedclass

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(C) Given velocity $\vec{v} = v_x \hat{i} + v_y \hat{j} = a \hat{i} + bx \hat{j}$.Thus,$v_x = \frac{dx}{dt} = a$ and $v_y = \frac{dy}{dt} = bx$.From $

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$y = \frac{bx^2}{2a}$

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(C) Given velocity $\vec{v} = v_x \hat{i} + v_y \hat{j} = a \hat{i} + bx \hat{j}$.Thus,$v_x = \frac{dx}{dt} = a$ and $v_y = \frac{dy}{dt} = bx$.From $v_x = a$,we integrate with respect to time: $\int_0^x dx = \int_0^t a dt$,which gives $x = at$,or $t = \frac{x}{a}$.Now,substitute $t$ into the expression for $v_y$: $\frac{dy}{dt} = bx$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{bx}{a}$.Integrating both sides with respect to $x$: $\int_0^y dy = \int_0^x \frac{b}{a} x dx$.This yields $y = \frac{b}{a} \cdot \frac{x^2}{2} = \frac{bx^2}{2a}$.

$A$ particle moves in the $x-y$ plane with velocity $\vec{v} = a\hat{i} + bx\hat{j}$,where $a$ and $b$ are constants. If the particle was initially at the origin,find the equation of its trajectory.

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