Chapter 5, Problem 65A

To determine

The distance between the ship and the destination. Also, the direction followed by ship to reach the destination.

Answer to Problem 65A

The distance between the ship and the destination is $509.9\text{km}$

The ship has to travel $78.69°$ south west to reach the destination.

Explanation of Solution

Given:

The initial distance between the ship and the port due south is $500.0\text{km}$ .

The wind blows the ship 100 km due east from its original position.

Formula used:

The resultant of two perpendicular vectors is,

  $R=\sqrt{P^2+Q^2}$

Where, P and Q are two vectors.

Calculation:

For the given situation consider the figure as shown below.

  

Consider the initial position of the ship is denoted by pointO and the position of the port is denoted by P.

The initial distance between the ship and the port is $OP=500.0\text{km}$ .

The ship is moved from position O to position N by the wind, such that $ON=100.0\text{km}$ .

The shortest distance between the new position of the ship and the port is denoted by NP.

Find the distance between the new position of the ship and the destination (port) as follows.

Apply Pythagoras theorem in $\Delta NOP$ .

  $\begin{array}{l}N{P}^2=O{P}^2+O{N}^2\\ NP=\sqrt{O{P}^2+O{N}^2}\\ NP=\sqrt{{\left(500.0\right)}^2+{\left(100.0\right)}^2}\\ NP=509.9\text{km}\end{array}$

Hence, the distance between the new position of the ship and the port is $NP=509.9\text{km}$ .

Consider the ship has to sail $NP=509.9\text{km}$ at an $\theta$ in the south west direction.

The angle $\theta$ is,

  $\begin{array}{l}\tan \theta =\left(\frac{OP}{ON}\right)\\ \theta ={\tan }^{-1}\left(\frac{500.0}{100.0}\right)\\ \theta =78.69°\end{array}$

Conclusion:

Therefore, the ship has to sail $NP=509.9\text{km}$ at an $78.69°$ in the south west direction.