JEE PYQ: Motion in a Plane - Question ID f5d69e69ed4e (JEE Main 2025)

In two-dimensional motion, the position vector $\overrightarrow{\mathrm{r}}(t)$ describes the location of a particle at any time $t$. The velocity vector $\overrightarrow{\mathrm{v}}(t)$ is the time derivative of the position vector: $\overrightarrow{\mathrm{v}}(t) = \frac{d\overrightarrow{\mathrm{r}}}{dt}.$ Its magnitude gives the speed, and its direction is tangent to the path at that instant.

If $\overrightarrow{\mathrm{r}} = x(t)\,\hat{i} + y(t)\,\hat{j}$, then $\overrightarrow{\mathrm{v}} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j}.$ The magnitude of $\overrightarrow{\mathrm{v}}$ is $|\overrightarrow{\mathrm{v}}| = \sqrt{\Bigl(\frac{dx}{dt}\Bigr)^2 + \Bigl(\frac{dy}{dt}\Bigr)^2},$ and its direction is specified by the angle $\theta$ it makes with one of the coordinate axes, where $\tan\theta = \frac{v_y}{v_x}.$

1. Identify the position components
Given $\overrightarrow{\mathrm{r}}(t) = \bigl(5\,t^2\bigr)\,\hat{i} + \bigl(-5\,t\bigr)\,\hat{j}\quad\text{(in meters)}.$ Thus $x(t) = 5\,t^2,\qquad y(t) = -5\,t.$

2. Compute the velocity components
Differentiate each component with respect to $t$: $v_x = \frac{dx}{dt} = \frac{d}{dt}\bigl(5\,t^2\bigr) = 10\,t,$ $v_y = \frac{dy}{dt} = \frac{d}{dt}\bigl(-5\,t\bigr) = -5.$ So the velocity vector at any time $t$ is $\overrightarrow{\mathrm{v}}(t) = 10\,t\,\hat{i} - 5\,\hat{j}.$

3. Evaluate at $t=2$ s
Substitute $t=2$: $v_x = 10\cdot 2 = 20,\qquad v_y = -5.$ Hence $\overrightarrow{\mathrm{v}}(2) = 20\,\hat{i} - 5\,\hat{j}.$

4. Find the magnitude of velocity
$|\overrightarrow{\mathrm{v}}| = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = \sqrt{25\cdot 17} = 5\sqrt{17}.$

5. Determine the direction
We compute the angle $\theta$ that $\overrightarrow{\mathrm{v}}$ makes with the negative $Y$–axis. Since $v_x=20$ points along $+\hat{i}$ and $v_y=-5$ points along $-\hat{j}$, the vector lies in the fourth quadrant. The angle between $\overrightarrow{\mathrm{v}}$ and the downward (negative $Y$) direction satisfies $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{|v_x|}{|v_y|} = \frac{20}{5} = 4.$ Therefore $\theta = \tan^{-1}4,$ measured from the negative $Y$–axis toward the positive $X$–axis.

6. Match with the given options
Our result is $\text{Magnitude} = 5\sqrt{17}\;\mathrm{m/s},\quad\text{Direction} = \tan^{-1}4\;\text{with the negative }Y\text{–axis}.$ This exactly matches option A.

1. Sign errors in differentiation: Students sometimes forget the negative sign when differentiating $-5t$, leading to $v_y=+5$ instead of $-5$. 2. Angle reference: One must be careful whether the angle is measured from the $X$–axis or the $Y$–axis. Here the question specifies “with –ve $Y$ axis,” so the tangent ratio uses $|v_x|/|v_y|$. 3. Magnitude simplification: $\sqrt{425}$ must be simplified to $5\sqrt{17}$; failing to do so can cause confusion when matching options.

Exam Tip: Always write down the velocity components explicitly, compute the magnitude, and then draw a small sketch of the vector to identify the correct reference axis for the angle.