Problem 83
Question
A man can swim with a speed of \(4 \mathrm{~km} / \mathrm{h}\) in still water. How long does he take to cross a river \(1 \mathrm{~km}\) wide, if the river flows steadily \(3 \mathrm{~km} / \mathrm{h}\) and he makes his strokes normal to the river current. How far down the river does he go when he reaches the other bank? [NCERT] (a) \(850 \mathrm{~m}\) (b) \(750 \mathrm{~m}\) (c) \(650 \mathrm{~m}\) (d) None of these
Step-by-Step Solution
Verified Answer
Option (b) 750 m.
1Step 1: Analyze Given Data
The man swims perpendicular to the river current at a speed of \(4 \text{ km/h}\) in still water. The width of the river is \(1 \text{ km}\) and the river's flow speed is \(3 \text{ km/h}\). We need to find the time taken to cross the river and how far he drifts downstream.
2Step 2: Determine Time to Cross the River
Since the man swims perpendicular to the current, his speed relative to the river crossing direction is his swimming speed, \(4\text{ km/h}\). To find the time, use the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). Here, the distance is \(1 \text{ km}\). Therefore, \( \text{Time} = \frac{1}{4} \text{ hours} = 0.25 \text{ hours} \).
3Step 3: Determine Downstream Distance
While crossing, the river current will carry him downstream. The speed of the river current is \(3\text{ km/h}\), and he takes \(0.25\text{ hours}\) to cross the river. The downstream distance is calculated by \( \text{Distance} = \text{Speed} \times \text{Time} \), so he will go \(3 \times 0.25 = 0.75 \text{ km}\) downstream which is \(750 \text{ m}\).
4Step 4: Concluding Answer
The downstream distance is \(750 \text{ m}\), which corresponds to option (b).
Key Concepts
Relative VelocitySwimming in CurrentTime and Distance CalculationsPhysics Problem Solving
Relative Velocity
Understanding relative velocity is crucial in this type of physics problem. Relative velocity refers to the velocity of an object as observed from a particular frame of reference. In the context of a river crossing problem, it involves evaluating the swimmer's speed in contrast to the river's current. Here, it's essential to determine how fast the man is moving relative to the riverbanks.
- The swimmer's velocity in still water is given as 4 km/h.
- The river has its own flow velocity of 3 km/h.
Swimming in Current
When a swimmer crosses a river with a current, it's important to consider the effect of the current on their trajectory. In this exercise, the swimmer is moving perpendicular to the current, which means:
- The swimmer's straightforward speed across the river remains 4 km/h because he maintains his strokes normal to the river's flow.
- The river's current will divert his path downstream by 3 km/h.
Time and Distance Calculations
Calculating time and distance effectively is a key part of solving river crossing problems. Here, the goal was to calculate how long it takes for the swimmer to cross the river and how far he is carried downstream.
To compute the crossing time, we use the swimmer's velocity perpendicular to the current, which is 4 km/h. The river's width is 1 km, and thus the time taken to cross is:
To compute the crossing time, we use the swimmer's velocity perpendicular to the current, which is 4 km/h. The river's width is 1 km, and thus the time taken to cross is:
- Time = Distance / Speed
- Time = 1 km / 4 km/h = 0.25 hours
- Downstream Distance = Speed of River x Time
- Downstream Distance = 3 km/h x 0.25 hours = 0.75 km = 750 meters
Physics Problem Solving
Physics problems like this one involve applying theoretical concepts to real-world scenarios. Here, abstract ideas become tangible through clear, step-by-step problem-solving processes. This task teaches us about:
- Decomposing vectors - Separating the swimmer’s velocity into components: one across the river and the other downstream with the current.
- Practical applications of formulas - Using time and distance equations to make accurate predictions.
- The balance of forces and motion - Understanding how natural forces, such as a river current, influence the path of an object.
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