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(C) Given velocity components are $v_x = \frac{dx}{dt} = 8t - 2$ and $v_y = \frac{dy}{dt} = 2$.Integrating $v_x$ with respect to $t$: $x = \int (8t -

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$x = y^2 - y + 2$

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(C) Given velocity components are $v_x = \frac{dx}{dt} = 8t - 2$ and $v_y = \frac{dy}{dt} = 2$.Integrating $v_x$ with respect to $t$: $x = \int (8t - 2) dt = 4t^2 - 2t + C_1$.At $t = 2 \, s$,$x = 14$,so $14 = 4(2)^2 - 2(2) + C_1 \Rightarrow 14 = 16 - 4 + C_1 \Rightarrow C_1 = 2$. Thus,$x = 4t^2 - 2t + 2$.Integrating $v_y$ with respect to $t$: $y = \int 2 dt = 2t + C_2$.At $t = 2 \, s$,$y = 4$,so $4 = 2(2) + C_2 \Rightarrow C_2 = 0$. Thus,$y = 2t$,which gives $t = \frac{y}{2}$.Substituting $t = \frac{y}{2}$ into the equation for $x$: $x = 4(\frac{y}{2})^2 - 2(\frac{y}{2}) + 2 = 4(\frac{y^2}{4}) - y + 2 = y^2 - y + 2$.

$A$ particle moves in the $xy$-plane with velocity $v_x = 8t - 2$ and $v_y = 2$. If it passes through the point $(14, 4)$ at $t = 2 \, s$,the equation of its path is:

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