JEE PYQ: Motion in a Plane - Question ID 9ef6e126bcdd (JEE Main 2024)

Core Formula & Concept:

In kinematics, the position vector $\vec{S}(t)$ of a particle moving in a plane (here the $Y$–$Z$ plane) is given as a function of time. The velocity $\vec{v}(t)$ is the time derivative of the position vector: $\vec{v}(t) = \frac{d\vec{S}}{dt}.$ Once $\vec{v}(t)$ is found, its magnitude is $|\vec{v}(t)| = \sqrt{v_y^2 + v_z^2},$ and its direction is along the unit vector $\hat{v} = \frac{\vec{v}}{|\vec{v}|}.$ In this problem the position is $\vec{S}(t)=2t^2\,\hat{j}+5\,\hat{k}$, so we differentiate each component with respect to $t$ to obtain the velocity components.

Step-by-Step Derivation:

1. Write the given position vector:
$\vec{S}(t) = 2\,t^2\,\hat{j} + 5\,\hat{k}.$

2. Compute the velocity vector by differentiating $\vec{S}(t)$ with respect to $t$:
$\vec{v}(t) = \frac{d\vec{S}}{dt} = \frac{d}{dt}\bigl(2t^2\bigr)\,\hat{j} + \frac{d}{dt}\bigl(5\bigr)\,\hat{k} = 4t\,\hat{j} + 0\,\hat{k} = 4t\,\hat{j}.$

3. Evaluate the velocity at $t=1$ s:
$\vec{v}(1) = 4\cdot1\,\hat{j} = 4\,\hat{j}\quad\text{(m/s)}.$

4. Determine the magnitude and direction of $\vec{v}(1)$:
- Magnitude: $|\vec{v}(1)| = \sqrt{4^2 + 0^2} = 4$ m/s. - Direction: purely along the $y$–axis (i.e.\ the $\hat{j}$ direction).

5. Match with the given options:
Option D reads “$4$ m/s in $y$–direction,” which exactly matches our result.

Common Traps & Exam Tip:

1. Forgetting to differentiate the constant term: The $5\,\hat{k}$ term is constant in time, so its derivative is zero. Students sometimes mistakenly write a nonzero $z$–component of velocity. 2. Misidentifying the plane: The problem states motion in the $Y$–$Z$ plane, but some may confuse axes and look for an $x$–component (which does not exist here). 3. Incorrect magnitude formula: One must use $\sqrt{v_y^2 + v_z^2}$, not simply add the components.

Exam Tip: Always check that the units of the answer match (here m/s) and that the direction corresponds to the correct axis.