JEE PYQ: Motion in a Plane - Question ID 8bb797b2d294 (JEE Main 2019)

ID: 8bb797b2d294JEE Main 2019Single Correct MCQ
The stream of a river is flowing with a speed of 2km/h. A swimmer can swim at a speed of 4km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?
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Step-by-step Explanation

Core Formula & Concept:

To solve this problem, we analyze the motion of the swimmer in two dimensions—along the river (downstream) and perpendicular to it (across the river). The key concept is relative velocity in a plane. The swimmer’s velocity relative to the water (vsw\vec{v}_{sw}) and the river’s velocity (vr\vec{v}_{r}) combine vectorially to give the swimmer’s resultant velocity relative to the ground (vsg\vec{v}_{sg}).

The condition to "cross the river straight" means the swimmer’s resultant velocity must be purely perpendicular to the river banks—i.e., no component along the river. Mathematically, this requires:

vsg=vsw+vr\vec{v}_{sg} = \vec{v}_{sw} + \vec{v}_{r}

For the resultant to have zero component along the river, the swimmer must swim at an angle θ\theta such that the component of vsw\vec{v}_{sw} along the river exactly cancels vr\vec{v}_{r}.

Step-by-Step Derivation:

1. Define Directions: Let the river flow along the positive x-axis. The swimmer aims to cross straight along the positive y-axis. Thus, the river velocity is: vr=2 km/h i^\vec{v}_{r} = 2 \text{ km/h} \ \hat{i} The swimmer’s velocity relative to water is: vsw=4 km/h (cosθ i^+sinθ j^)\vec{v}_{sw} = 4 \text{ km/h} \ (\cos\theta \ \hat{i} + \sin\theta \ \hat{j}) where θ\theta is the angle the swimmer makes with the river flow (x-axis).

2. Resultant Velocity Condition: The swimmer’s resultant velocity relative to ground is: vsg=vsw+vr=(4cosθ+2)i^+4sinθ j^\vec{v}_{sg} = \vec{v}_{sw} + \vec{v}_{r} = (4\cos\theta + 2)\hat{i} + 4\sin\theta \ \hat{j} For crossing straight, the x-component must vanish: 4cosθ+2=04\cos\theta + 2 = 0

3. Solve for θ\theta: 4cosθ=24\cos\theta = -2 cosθ=12\cos\theta = -\frac{1}{2} The angle whose cosine is 12-\frac{1}{2} in the range 0°θ180°0° \leq \theta \leq 180° is: θ=120°\theta = 120°

4. Interpretation: The swimmer must swim at 120°120° with respect to the river flow (x-axis) to cancel the downstream drift and cross straight.

Common Traps & Exam Tip:

- Misinterpreting the angle: Students often confuse whether the angle is measured with respect to the river flow or the perpendicular. Always define the reference direction clearly. - Sign errors in components: Forgetting that the swimmer’s x-component must oppose the river flow leads to incorrect angles. Double-check the vector addition. - Unit consistency: Ensure all speeds are in the same units (km/h here) to avoid calculation errors.

Exam Tip: Draw a clear velocity vector diagram. The swimmer’s velocity vector should form an obtuse angle with the river flow to counteract the downstream current.