JEE PYQ: Motion in a Plane - Question ID 6e70010b4082 (JEE Main 2024)

Core Formula & Concept:

In kinematics, the motion of a particle in a plane is described by its position coordinates as functions of time, $x(t)$ and $y(t)$. To analyze the nature of the motion, we use the following key concepts and formulas:

• Velocity: The velocity components are the first time derivatives of the position coordinates: $v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}$
• Acceleration: The acceleration components are the first time derivatives of the velocity components (or second derivatives of position): $a_x = \frac{dv_x}{dt} = \frac{d^2 x}{dt^2}, \quad a_y = \frac{dv_y}{dt} = \frac{d^2 y}{dt^2}$
• Uniform vs Non-Uniform Acceleration:
  • If both $a_x$ and $a_y$ are constant (i.e., do not depend on time), the motion is uniformly accelerated.
  • If either $a_x$ or $a_y$ varies with time, the motion is non-uniformly accelerated.
• Path of Motion:
  • If the path is a straight line, the ratio $\frac{y}{x}$ or $\frac{dy}{dx}$ is constant.
  • If $y$ is a quadratic function of $x$, the path is typically parabolic.

In this problem, we are given: $x(t) = 2 + 4t, \quad y(t) = 3t + 8t^2$ We will compute velocity and acceleration components, check their constancy, and analyze the path to determine the correct option.

--- Step-by-Step Derivation:

Step 1: Compute velocity components

$v_x = \frac{dx}{dt} = \frac{d}{dt}(2 + 4t) = 4$ $v_y = \frac{dy}{dt} = \frac{d}{dt}(3t + 8t^2) = 3 + 16t$

So, $v_x = 4$ (constant), and $v_y = 3 + 16t$ (varies linearly with time).

Step 2: Compute acceleration components

$a_x = \frac{dv_x}{dt} = \frac{d}{dt}(4) = 0$ $a_y = \frac{dv_y}{dt} = \frac{d}{dt}(3 + 16t) = 16$

Thus, $a_x = 0$ and $a_y = 16$, both of which are constant. This means the acceleration is uniform.

Step 3: Analyze the path of motion

To find the path, we eliminate the parameter $t$ from $x(t)$ and $y(t)$.

From $x = 2 + 4t$, solve for $t$: $t = \frac{x - 2}{4}$ Substitute into $y$: $y = 3t + 8t^2 = 3\left(\frac{x - 2}{4}\right) + 8\left(\frac{x - 2}{4}\right)^2$ Simplify: $y = \frac{3}{4}(x - 2) + 8 \cdot \frac{(x - 2)^2}{16} = \frac{3}{4}(x - 2) + \frac{1}{2}(x - 2)^2$ This is a quadratic equation in $x$: $y = \frac{1}{2}(x - 2)^2 + \frac{3}{4}(x - 2)$

A quadratic relationship between $y$ and $x$ indicates a parabolic path.

Step 4: Match with given options

• A: Uniform motion along a straight line — Incorrect. Velocity is not constant (since $v_y$ depends on $t$), and path is not straight.
• B: Non-uniformly accelerated — Incorrect. Acceleration is constant (uniform).
• C: Uniformly accelerated having motion along a straight line — Incorrect. Path is parabolic, not straight.
• D: Uniformly accelerated having motion along a parabolic path — Correct. Matches our findings.

--- Common Traps & Exam Tip:

Trap 1: Students often confuse constant velocity with constant acceleration. Here, $v_x$ is constant, but $v_y$ is not — so velocity is not constant. However, acceleration is constant. This leads some to choose option A (uniform motion), which is wrong.

Trap 2: Another common mistake is assuming that if one component of acceleration is zero ($a_x = 0$), the motion is not uniformly accelerated. But uniform acceleration only requires that both $a_x$ and $a_y$ are constant — they can be zero or non-zero. Here, $a_x = 0$, $a_y = 16$, both constant → uniformly accelerated.

Trap 3: Students may check only the acceleration and conclude option C (uniformly accelerated along straight line), forgetting to analyze the path. The path is crucial — it's parabolic, not straight.

Exam Tip: Always:

• Compute both velocity and acceleration components.
• Check if acceleration is constant (uniform).
• Eliminate time to find the path equation.
• Match all aspects (acceleration type and path shape) with the options.

In this case, the motion is uniformly accelerated and follows a parabolic path → Option D is correct.