The Question is from neet-chemistry atomic-structure motion-of-a-boat-in-river, and it was created by the author P SRAVAN KUMAR

Question

A swimmer can swim with a velocity of $v$ m/s in still water. He wants to cross a river of width $d$ meters flowing with a velocity $u$ m/s.  If the swimmer heads in a direction making an angle $	heta$ with the upstream direction such that he crosses the river in minimum time, what is the drift experienced by the swimmer?

P SRAVAN KUMAR · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 1
## CONCEPT FRAMEWORK
 **Primary Concept**: Relative velocity and minimum time to cross a river.
 **Related Topics**: Vectors, trigonometry, kinematics.
 **Prerequisites**: Understanding of vector addition and resolution of velocities.
## SOLUTION PATHWAY
**Given Information**:
  * Swimmer's velocity in still water: $v$ m/s
  * River's velocity: $u$ m/s
  * Width of the river: $d$ meters
  * Swimmer's direction is at angle θ with upstream direction.
  * Minimum time to cross the river is required.
**Method & Steps**:
 1. **Resolve the swimmer's velocity:** The swimmer's velocity ($v$) can be resolved into two components: one perpendicular to the river flow ($v_y = v\cos	heta$) and one parallel to the river flow ($v_x = v\sin	heta$).
  Reasoning: To minimize the time, the swimmer should swim perpendicular to the river bank.  The component of the swimmer's velocity perpendicular to the flow determines the time to cross.
  Formula/Rule: Vector resolution
 2. **Find the time to cross:** The time taken to cross the river is given by the distance divided by the velocity component perpendicular to the flow: $t = \frac{d}{v\cos	heta}$. To minimize time, $\cos	heta$ must be maximum, which occurs when $	heta = 0$. This means the swimmer should swim directly across the river (perpendicular to the flow).
  Working:  If the swimmer swims directly across, the time taken is $t_{min} = \frac{d}{v}$.
  Key Point: Minimum time is achieved when the swimmer swims perpendicular to the river bank.
 3. **Calculate the drift:** When the swimmer swims directly across, the only velocity component in the direction of the river flow is the river's velocity ($u$). The drift is the distance covered in the direction of the river flow during the crossing time. 
  Result: Drift = $u 	imes t_{min} = u 	imes \frac{d}{v} = d\frac{u}{v}$
  Verification: The units are meters, which is consistent with a distance. The formula shows that the drift is directly proportional to the river velocity and inversely proportional to the swimmer's velocity in still water.
 ## OPTION ANALYSIS
[A] ✓: $d \frac{u}{v}$: This is the correct answer, as derived above.  The drift is the product of the river's velocity and the time taken to cross.
[B] :x: $d \frac{v}{u}$: This is incorrect. It inverts the relationship between the swimmer's and river's velocities.
[C] :x: $d \sqrt{\frac{u^2}{v^2} + 1}$: This formula is incorrect and doesn't account for the minimum time condition.
[D] :x: $0$: This is incorrect. There will always be a drift unless the river velocity is zero.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. [Error 1]: Assuming the swimmer swims directly across the river and ignoring the river's current.  [How to avoid]: Always consider the effect of the river current on the swimmer's path.
  2. [Error 2]: Not resolving the swimmer's velocity into components. [How to avoid]: Draw a vector diagram to resolve the swimmer's velocity into components parallel and perpendicular to the river flow.
 **Quick Tips**:
  1. [Helpful shortcut]: For minimum time, the swimmer should swim perpendicular to the river bank.
  2. [Memory aid]: Remember that drift is the product of river velocity and crossing time.

P SRAVAN KUMAR · Answer

The correct Option is "$d \frac{u}{v}$"

P SRAVAN KUMAR · Answer

$d \frac{v}{u}$

P SRAVAN KUMAR · Answer

$d \sqrt{\frac{u^2}{v^2} + 1}$

P SRAVAN KUMAR · Answer

$0$

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