QJEEM24Phf5cf: JEE Main 2024 Physics - Class 11 JEEM24Phf5cf Solution: The co-ordinates of a particle moving in - plane are given by : …

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 $t$ :

• $x = 2 + 4t$
• $y = 3t + 8t^2$

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.

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.

The equation for $y$ can show two types of motion depending on the terms:

1. If it was of the form $y = y_0 + vt$ , it would indicate uniform motion.
2. 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.

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.

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.

Hence, the correct answer is:

Option D: uniformly accelerated having motion along a parabolic path.