The Question is from NEET guide , and it was created by the author Mr. S M HUSSAIN

Question

A boat crosses a river from port A to port B, which are just on the opposite side. The speed of the water is ${V_W}$ and that of boat is ${V_B}$ relative to still water. Assume ${V_B} = 2{V_W}$. What is the time taken by the boat, if It has to cross the river directly on the AB line

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Relative Velocity
	**Related Topics**: Vector addition, Trigonometry, Time calculation
	**Prerequisites**: Understanding of vectors, basic trigonometry, and the concept of relative motion.
## SOLUTION PATHWAY
**Given Information**:
	* Speed of water: $V_W$
	* Speed of boat in still water: $V_B = 2V_W$
	* Distance across the river: D
	* Boat needs to travel directly across the river (along the AB line).
**Method & Steps**:
	1.  **Determine the boat's velocity relative to the ground:**
		Reasoning: The boat must point upstream to cancel the water's flow and move directly across the river. Let $	heta$ be the angle between the boat's velocity vector and the line perpendicular to the river flow. The boat's velocity relative to the ground should be perpendicular to the river flow.
		Formula/Rule: Vector addition and trigonometry.
	2.  **Calculate the angle $	heta$:**
		Working: The component of the boat's velocity along the river flow should cancel the water's velocity. So, $V_B \sin(	heta) = V_W$. Since $V_B = 2V_W$, we have $2V_W \sin(	heta) = V_W$, which simplifies to $\sin(	heta) = 1/2$. Therefore, $	heta = 30^\circ$.
		Key Point:  The boat's velocity vector is at an angle of 30 degrees upstream relative to the direction perpendicular to the river flow.
	3.  **Calculate the boat's effective velocity across the river:**
		Working: The component of the boat's velocity perpendicular to the river flow is $V_B \cos(	heta) = V_B \cos(30^\circ) = V_B \frac{\sqrt{3}}{2}$.
		Key Point: This is the actual speed of the boat in the direction across the river.
	4.  **Calculate the time taken to cross the river:**
		Working: Time = Distance / Speed. The distance is D, and the effective speed across the river is $V_B \frac{\sqrt{3}}{2}$. So, Time = $D / (V_B \frac{\sqrt{3}}{2}) = \frac{2D}{V_B \sqrt{3}}$.
		Result: The time taken is $\frac{2D}{V_B \sqrt{3}}$.
		Verification: The calculated time matches the correct answer.
 ## OPTION ANALYSIS
[A] ✓: The boat's effective speed across the river is $V_B\cos(30) = V_B \frac{\sqrt{3}}{2}$.  Time = Distance/Speed = $D/(V_B \frac{\sqrt{3}}{2}) = \frac{2D}{V_B\sqrt{3}}$
[B] :x: This is the inverse of the correct answer.
[C] :x: This doesn't consider the correct angle or the boat's velocity components.
[D] :x: This is not the right formula for calculating the time.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrectly resolving vectors:** Errors in resolving the boat's velocity into components parallel and perpendicular to the river flow can lead to wrong results.  Carefully consider the angles.
		2. **Forgetting relative motion:** Failing to account for the water's velocity will lead to wrong calculations. Remember that the boat's velocity relative to the ground is what matters for crossing the river directly.
	**Quick Tips**:
		1. **Draw a diagram:** A clear diagram of the vectors involved can help in understanding the problem.
		2. **Break into components:** Resolve velocities into components to simplify calculations.

Mr. S M HUSSAIN · Answer

The correct Option is "$\frac{{2D}}{{{V_B}\sqrt 3 }}$"

Mr. S M HUSSAIN · Answer

$\frac{{\sqrt 3 D}}{{2{V_B}}}$

Mr. S M HUSSAIN · Answer

$\frac{D}{{{V_B}\sqrt 2 }}$

Mr. S M HUSSAIN · Answer

$\frac{{D\sqrt 2 }}{{{V_B}}}$

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