Chapter 1, Problem 10AP

A sailboat heads north at 3 m/s for 1 hour and then tacks back to the southeast (at 45° to north) at 2 m/s for 45 minutes.

a. How far has the boat sailed?

b. How far is it from its starting location?

Summary Introduction

To find: The distance sailed by the boat in north and south east direction.

Answer to Problem 10AP

The total distance sailed in north and southeast direction is $16.2\text{km}$.

Explanation of Solution

Calculation:

Write an expression to find the distance sailed by the boat in north direction.

$\begin{array}{l}\text{Distance}\text{sailed}\\ \text{by}\text{the}\text{boat}\text{towards}\text{north}\end{array}\}={\left(\text{average}\text{speed}\times \text{time}\right)}_{\text{north}}$

Substitute $3\text{m}/\text{s}$ for average speed and 1 hr for time to find the distance sailed by the boat in north direction.

$\begin{array}{c}\begin{array}{l}\text{Distance}\text{sailed}\\ \text{by}\text{the}\text{boat}\text{towards}\text{north}\end{array}\}={\left(\left(\text{3}\text{m}/\text{s}\right)\left(1\text{hr}\right)\right)}_{\text{north}}\\ =\text{3}\frac{\text{m}}{\text{s}}\left(\frac{\text{1}\text{km}}{\text{1,000}\text{m}}\right)\left(1\text{hr}\frac{3,600\text{s}}{1\text{hr}}\right)\\ =10.8\text{km}\end{array}$

Thus, distance sailed by the boat in north direction is $10.8\text{km}$.

Write an expression to find the distance sailed by the boat in southeast direction.

$\begin{array}{l}\text{Distance}\text{sailed}\text{by}\text{the}\text{boat}\\ \text{in}\text{southeast}\text{direction}\end{array}\}={\left(\text{average}\text{speed}\times \text{time}\right)}_{\text{southeast}}$

Substitute $2\text{m}/\text{s}$ for average speed and 45 min for time to find the distance sailed by the boat in southeast direction.

$\begin{array}{c}\begin{array}{l}\text{Distance}\text{sailed}\text{by}\text{the}\text{boat}\\ \text{in}\text{southeast}\text{direction}\end{array}\}={\left(\left(\text{2}\text{m}/\text{s}\right)\left(45\text{min}\right)\right)}_{\text{southeast}}\\ =\text{2}\frac{\text{m}}{\text{s}}\left(\frac{\text{1}\text{km}}{\text{1,000}\text{m}}\right)\left(45\text{min}\frac{60\text{s}}{1\text{min}}\right)\\ =5.4\text{km}\end{array}$

Thus, distance sailed by the boat in southeast direction is $5.4\text{km}$.

Write an expression to find the total distance sailed in north and southeast direction.

$\begin{array}{l}\text{Total}\text{distance}\text{sailed}\text{in}\text{north}\\ \text{and}\text{southeast}\text{direction}\end{array}\}=\{\begin{array}{l}\text{distance}\text{sailed}\text{in}\text{north}\\ +\text{distance}\text{sailed}\text{in}\text{southeast}\end{array}$

Substitute $10.8\text{km}$ for distance sailed in north and $5.4\text{km}$ for distance sailed in southeast to find the total distance.

$\begin{array}{c}\begin{array}{l}\text{Total}\text{distance}\text{sailed}\text{in}\text{north}\\ \text{and}\text{southeast}\text{direction}\end{array}\}=10.8\text{km}+5.4\text{km}\\ =16.2\text{km}\end{array}$

Conclusion

Therefore, the total distance sailed in north and southeast direction is $16.2\text{km}$.

Summary Introduction

To find: The distance of the boat from its starting location.

Answer to Problem 10AP

The distance of the boat from its starting location is $8\text{km}$.

Explanation of Solution

Calculation:

Write an expression to find the distance of the boat from its starting location.

$A^2=B^2+C^2-2BC\cos \theta$

Draw the diagram to illustrate the situation in this problem.

Substitute 10.8 km for B and 5.4 km for C to find the distance of the boat from its starting location.

$\begin{array}{c}{A}^2={\left(10.8\text{km}\right)}^2+{\left(5.4\text{km}\right)}^2-2\left(10.8\text{km}\right)\left(5.4\text{km}\right)\cos 45°\\ A=\sqrt{63.323{\text{km}}^2}\\ =7.957\text{km}\\ \approx 8\text{km}\end{array}$

Conclusion

Therefore, the distance of the boat from its starting location is $8\text{km}$.