JEE PYQ: Motion in a Plane - Question ID 189199d70862 (JEE Main 2021)

In problems involving relative motion in a plane (such as a swimmer crossing a river), the key concept is vector addition of velocities. The swimmer's velocity relative to the water ($\vec{v}_{sw}$) and the river's velocity relative to the ground ($\vec{v}_{rw}$) combine to give the swimmer's resultant velocity relative to the ground ($\vec{v}_{sg}$).

The fundamental relation is: $\vec{v}_{sg} = \vec{v}_{sw} + \vec{v}_{rw}$

For the swimmer to reach a point directly opposite on the other side of the river, the net horizontal displacement due to the river's flow must be zero. This means the horizontal component of the swimmer's velocity relative to the water must exactly cancel the river's velocity.

Key formulas used:

1. Components of a vector: If a vector $\vec{A}$ makes an angle $\theta$ with the positive x-axis, its components are: $A_x = A \cos \theta$ $A_y = A \sin \theta$
2. Condition for zero net horizontal displacement: $v_{sw,x} + v_{rw} = 0$ (where $v_{sw,x}$ is the horizontal component of the swimmer's velocity relative to water).

Step 1: Define the coordinate system
Let the river flow along the positive x-axis. The swimmer's velocity relative to the water ($\vec{v}_{sw}$) is given as 10 m/s at an angle of $120^\circ$ with the river flow (x-axis).

Step 2: Resolve the swimmer's velocity into components
The angle $120^\circ$ is measured from the positive x-axis (river flow direction). The components of $\vec{v}_{sw}$ are: $v_{sw,x} = v_{sw} \cos 120^\circ = 10 \cdot \cos 120^\circ$ $v_{sw,y} = v_{sw} \sin 120^\circ = 10 \cdot \sin 120^\circ$

Using trigonometric values: $\cos 120^\circ = -\frac{1}{2}, \quad \sin 120^\circ = \frac{\sqrt{3}}{2}$ Thus: $v_{sw,x} = 10 \cdot \left(-\frac{1}{2}\right) = -5 \text{ m/s}$ $v_{sw,y} = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m/s}$

Step 3: Apply the condition for zero net horizontal displacement
For the swimmer to reach the point directly opposite, the resultant horizontal velocity must be zero. The river's velocity ($\vec{v}_{rw}$) is along the positive x-axis with magnitude $x$ m/s. The net horizontal velocity is: $v_{sg,x} = v_{sw,x} + v_{rw} = -5 + x$ For zero net horizontal displacement: $-5 + x = 0$ Thus: $x = 5 \text{ m/s}$

Step 4: Verify the result
The swimmer's vertical component ($5\sqrt{3}$ m/s) ensures they cross the river, while the horizontal component ($-5$ m/s) cancels the river's flow ($5$ m/s), resulting in no net horizontal drift. This confirms the swimmer reaches the point directly opposite.

Conclusion: The value of $x$ to the nearest integer is 5.

1. Misinterpreting the angle: Students often confuse whether the angle is measured with respect to the river flow or the perpendicular. Here, $120^\circ$ is with respect to the river flow (x-axis), not the perpendicular (y-axis). Drawing a diagram is crucial.
2. Sign errors in components: The horizontal component of the swimmer's velocity is negative because it opposes the river flow. Forgetting the negative sign leads to $x = -5$, which is physically meaningless.
3. Ignoring the condition for zero drift: Some students assume the swimmer's resultant velocity must be perpendicular to the river, which is not the case here. The key is zero net horizontal displacement, not perpendicularity.
4. Trigonometric errors: Using $\cos 120^\circ = \frac{1}{2}$ (incorrect) instead of $-\frac{1}{2}$ is a common mistake. Always verify trigonometric values for angles greater than $90^\circ$.