### Problem StatementA swimmer is capable of swimming 0.47 m/s in still water.If she aims her body directly across a 77-m wide river whose current is 0.41 m/s, how far downstream (from a point opposite her starting point) will she land?\[ d = \begin{array}{|c|l|} \hline \text{value?} & \text{units?} \\ \hline \end{array} \]How long will it take her to reach the other side?\[ t = \begin{array}{|c|l|} \hline \text{value?} & \text{units?} \\ \hline \end{array} \]### Explanation:Here, we have a scenario where a swimmer is crossing a river with a given current speed. The swimmer's speed is perpendicular to the current of the river. The diagram in the example could be interpreted as showing the swimmer's intended path directly across the river and the river's current affecting the swimmer's actual path.**Key Elements:**1. **Swimmer's speed in still water:** 0.47 m/s2. **River width:** 77 meters3. **River current speed:** 0.41 m/sThe questions require us to compute the following:- **Downstream distance**: How far downstream the river current will carry the swimmer.- **Travel time**: How long it will take the swimmer to cross the river.To solve this problem, we can use the following principles:1. **Time to cross the river (t):** - Using the swimmer's speed in still water and the width of the river, we will compute the time \( t \). \[ t = \frac{\text{distance across the river}}{\text{swimmer's speed}} = \frac{77 \text{ m}}{0.47 \text{ m/s}} \]2. **Downstream distance (d):** - Using the river current speed and the time \( t \), we compute how far downstream the swimmer will be carried. \[ d = \text{river current speed} \times t = 0.41 \text{ m/s} \times t \]The solution requires substituting \( t \) from the first equation into the second to find \( d

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A Swimmer 1s Capable oe Soimming 0.47 m/S in Stiu Water If She cams her bady drectly acrss a 77-m Wide River whose current is Ô.41 M/S, how for downstream from a point oppIste her stating Point)will she land> vatre ? /I inits?

A Swimmer 1s Capable oe Soimming 0.47 m/S in Stiu Water If She cams her bady drectly acrss a 77-m Wide River whose current is Ô.41 M/S, how for downstream from a point oppIste her stating Point)will she land> vatre ? /I inits?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Density of Solids

The density of a substance refers to how much matter is present in a given amount of space. It is expressed as the ratio of the mass of the substance to its volume.

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### Problem Statement

A swimmer is capable of swimming 0.47 m/s in still water.

If she aims her body directly across a 77-m wide river whose current is 0.41 m/s, how far downstream (from a point opposite her starting point) will she land?

\[ d = \begin{array}{|c|l|} \hline \text{value?} & \text{units?} \\ \hline \end{array} \]

How long will it take her to reach the other side?

\[ t = \begin{array}{|c|l|} \hline \text{value?} & \text{units?} \\ \hline \end{array} \]

### Explanation:

Here, we have a scenario where a swimmer is crossing a river with a given current speed. The swimmer's speed is perpendicular to the current of the river. The diagram in the example could be interpreted as showing the swimmer's intended path directly across the river and the river's current affecting the swimmer's actual path.

**Key Elements:**

1. **Swimmer's speed in still water:** 0.47 m/s
2. **River width:** 77 meters
3. **River current speed:** 0.41 m/s

The questions require us to compute the following:

- **Downstream distance**: How far downstream the river current will carry the swimmer.
- **Travel time**: How long it will take the swimmer to cross the river.

To solve this problem, we can use the following principles:

1. **Time to cross the river (t):**
- Using the swimmer's speed in still water and the width of the river, we will compute the time \( t \).

\[
t = \frac{\text{distance across the river}}{\text{swimmer's speed}} = \frac{77 \text{ m}}{0.47 \text{ m/s}}
\]

2. **Downstream distance (d):**
- Using the river current speed and the time \( t \), we compute how far downstream the swimmer will be carried.

\[
d = \text{river current speed} \times t = 0.41 \text{ m/s} \times t
\]

The solution requires substituting \( t \) from the first equation into the second to find \( d

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