JEE PYQ: Motion in a Plane - Question ID a394579452e6 (JEE Main 2021)

To determine the direction of the mosquito at any instant, we must analyze its velocity vector at that moment. The velocity vector $\overrightarrow{v}(t)$ is given as a function of time: $\overrightarrow{v}(t) = 0.5\,t^2\,\widehat{i} + 3t\,\widehat{j} + 9\,\widehat{k}\quad\text{(m/s)}$

The direction of motion is defined by the unit vector in the direction of $\overrightarrow{v}(t)$. However, since the question asks for the direction relative to a coordinate axis (either the x-axis or y-axis), we focus on the angle that the velocity vector makes with one of these axes.

The angle $\theta$ between the velocity vector and a chosen axis (say, the y-axis) can be found using the tangent of the angle between the projection of $\overrightarrow{v}$ in the plane perpendicular to that axis and the component along the axis itself. Specifically:

• If we want the angle from the y-axis, we consider the components in the $x$-$z$ plane (perpendicular to $y$) and the $y$-component.
• The tangent of the angle from the y-axis is: $\tan\theta = \frac{\text{Magnitude of projection in } x\text{-}z \text{ plane}}{v_y}$

Step 1: Compute the velocity vector at $t = 2$ s

Substitute $t = 2$ into the given velocity expression: $$\overrightarrow{v}(2) = 0.5 \cdot (2)^2\,\widehat{i} + 3 \cdot 2\,\widehat{j} + 9\,\widehat{k} = 0.5 \cdot 4\,\widehat{i} + 6\,\widehat{j} + 9\,\widehat{k} = 2\,\widehat{i} + 6\,\widehat{j} + 9\,\widehat{k}$$ So, the velocity components at $t = 2$ s are: $v_x = 2,\quad v_y = 6,\quad v_z = 9$

Step 2: Determine the angle from the y-axis

The question asks for the direction from the y-axis. This means we want the angle $\theta$ between the velocity vector and the y-axis.

The y-axis is along $\widehat{j}$. The component of $\overrightarrow{v}$ along the y-axis is $v_y = 6$. The remaining components ($v_x$ and $v_z$) lie in the plane perpendicular to the y-axis (the $x$-$z$ plane).

The magnitude of the projection of $\overrightarrow{v}$ in the $x$-$z$ plane is: $\sqrt{v_x^2 + v_z^2} = \sqrt{2^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$

The tangent of the angle $\theta$ from the y-axis is the ratio of the perpendicular component (in the $x$-$z$ plane) to the parallel component (along $y$): $\tan\theta = \frac{\sqrt{v_x^2 + v_z^2}}{v_y} = \frac{\sqrt{85}}{6}$ Thus, the angle $\theta$ is: $\theta = \tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$

Step 3: Match with the given options

The derived expression matches Option A: $\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\quad\text{from y-axis}$

1. Misidentifying the reference axis: Students often confuse whether the angle is measured from the x-axis, y-axis, or z-axis. The question explicitly asks for the angle from the y-axis, so the perpendicular components are $v_x$ and $v_z$.
2. Incorrect projection: Some students mistakenly include $v_y$ in the numerator when calculating the angle from the y-axis. The numerator should only include components perpendicular to the reference axis.
3. Unit vector confusion: The question asks for direction, not the unit vector. While the unit vector gives the direction, the angle is derived from the components of $\overrightarrow{v}$ directly.
4. Arithmetic errors: Calculating $\sqrt{2^2 + 9^2}$ as $\sqrt{4 + 81} = \sqrt{85}$ is critical. A common mistake is to miscompute this as $\sqrt{89}$ or $\sqrt{77}$.

Exam Tip: Always sketch the velocity vector and label its components. This helps visualize which components are parallel and perpendicular to the reference axis.