JEE PYQ: Motion in a Plane - Question ID f4df7366282f (JEE Main 2021)

In problems involving motion in a plane, the key concept is vector addition of velocities. When two or more velocities act on a body simultaneously, the resultant velocity is obtained by adding the individual velocity vectors using the rules of vector algebra.

The fundamental formulas used are:

• Resolution of a vector into its components:
  If a velocity vector $\vec{v}$ makes an angle $\theta$ with the positive x-axis, its components are: $v_x = v \cos \theta$ $v_y = v \sin \theta$
• Resultant velocity vector:
  If two velocity vectors $\vec{v_1}$ and $\vec{v_2}$ act on a body, the resultant velocity $\vec{v_R}$ is: $\vec{v_R} = \vec{v_1} + \vec{v_2}$ Its magnitude is: $|\vec{v_R}| = \sqrt{v_{Rx}^2 + v_{Ry}^2}$
• Displacement from velocity:
  For constant velocity, displacement $\vec{s}$ in time $t$ is: $\vec{s} = \vec{v_R} \cdot t$

Step 1: Understand the directions and assign coordinate axes

Let’s define:

• Positive x-axis: East direction
• Positive y-axis: North direction

Step 2: Resolve the butterfly’s velocity into components

Given: Butterfly’s velocity $v_b = 4\sqrt{2}$ m/s in North-East direction.
Since North-East is at $45^\circ$ to both North and East, the components are: $v_{bx} = v_b \cos 45^\circ = 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4 \text{ m/s (East)}$ $v_{by} = v_b \sin 45^\circ = 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4 \text{ m/s (North)}$

Step 3: Resolve the wind velocity into components

Wind is blowing from North to South at 1 m/s.
This means the wind velocity vector points South, i.e., in the negative y-direction.
So, $v_{wx} = 0 \text{ m/s (no East-West component)}$ $v_{wy} = -1 \text{ m/s (South)}$

Step 4: Find the resultant velocity vector

The resultant velocity $\vec{v_R}$ is the vector sum of the butterfly’s velocity and the wind velocity: $v_{Rx} = v_{bx} + v_{wx} = 4 + 0 = 4 \text{ m/s}$ $v_{Ry} = v_{by} + v_{wy} = 4 + (-1) = 3 \text{ m/s}$

Step 5: Compute the magnitude of the resultant velocity

$|\vec{v_R}| = \sqrt{v_{Rx}^2 + v_{Ry}^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m/s}$

Step 6: Calculate the displacement in 3 seconds

Displacement $s = |\vec{v_R}| \cdot t = 5 \cdot 3 = 15 \text{ m}$

Step 7: Match with the given options

The resultant displacement is 15 m, which corresponds to option D.

Trap 1: Misinterpreting wind direction
Students often confuse "wind blowing from North to South" with "wind blowing toward North". The correct interpretation is that the wind velocity vector points South, so it should be assigned a negative y-component.

Trap 2: Incorrect angle for North-East
Some students mistakenly use $30^\circ$ or $60^\circ$ for North-East. North-East is exactly $45^\circ$ from both North and East.

Trap 3: Forgetting to compute displacement
The question asks for displacement, not velocity. Students may stop at finding the resultant velocity and forget to multiply by time.

Exam Tip: Always draw a quick sketch of the directions and label the components. This helps avoid sign errors and clarifies the vector addition.