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(D) Given the position coordinates as functions of time:$x = 2 + 4t$$y = 3t + 8t^2$First,find the velocity components by differentiating with respect

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uniformly accelerated having motion along a parabolic path.

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(D) Given the position coordinates as functions of time:$x = 2 + 4t$$y = 3t + 8t^2$First,find the velocity components by differentiating with respect to time $t$:$v_x = \frac{dx}{dt} = 4$$v_y = \frac{dy}{dt} = 3 + 16t$Next,find the acceleration components by differentiating the velocity components:$a_x = \frac{dv_x}{dt} = 0$$a_y = \frac{dv_y}{dt} = 16$Since the acceleration components are constants ($a_x = 0$ and $a_y = 16$),the particle undergoes uniformly accelerated motion.To determine the path,express $t$ in terms of $x$ from the first equation:$t = \frac{x - 2}{4}$Substitute this into the equation for $y$:$y = 3\left(\frac{x - 2}{4}\right) + 8\left(\frac{x - 2}{4}\right)^2$Since this equation is of the form $y = Ax^2 + Bx + C$,it represents a parabolic path.Therefore,the motion is uniformly accelerated along a parabolic path.

The coordinates of a particle moving in the $x-y$ plane are given by: $x = 2 + 4t$,$y = 3t + 8t^2$. The motion of the particle is:

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