The co-ordinates of a particle moving in $x$-$y$ plane are given by : $x=2+4 \mathrm{t}, y=3 \mathrm{t}+8 \mathrm{t}^2$.

The motion of the particle is :

<p>To determine the nature of the motion of the particle given by its coordinates in the $x$-$y$ plane, we analyze the given equations for $x$ and $y$ in terms of time $t$:</p> <ul> <li>$x = 2 + 4t$</li> <li>$y = 3t + 8t^2$</li> </ul> <p>Firstly, the equation for $x$ is of the form $x = x_0 + vt$, where $x_0 = 2$ is the initial position and $v = 4$ is the constant velocity along the $x$-axis. This suggests a uniform motion along the $x$-axis because the velocity remains constant with time.</p> <p>Secondly, the equation for $y$ is a second-degree polynomial in $t$, which indicates a parabolic path. The presence of the $t^2$ term ($8t^2$) signifies acceleration since the position along the $y$-axis is changing at a rate that itself changes over time.</p> <p>The equation for $y$ can show two types of motion depending on the terms: <ol> <li>If it was of the form $y = y_0 + vt$, it would indicate uniform motion.</li> <li>If it was of the form $y = y_0 + vt + \frac{1}{2}at^2$, where $a$ would represent acceleration, it would indicate uniformly accelerated motion. The presence of the $8t^2$ term here plays a similar role, indicating that the motion is uniformly accelerated in the $y$-direction due to the constant acceleration implied by this term.</li> </ol> </p> <p>Since the motion in the $y$ direction is determined by a quadratic equation, and the path of the particle depends on both the $x$ and $y$ coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the $y$-coordinate.</p> <p>Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the $t^2$ term in the $y$ equation, indicating uniform acceleration. Therefore, the motion is uniformly accelerated and follows a parabolic path.</p> <p>Hence, the correct answer is:</p> <p>Option D: uniformly accelerated having motion along a parabolic path.</p>