Chapter 9.3, Problem 54E

(a)

To determine

To find: The x -coordinate of the position of the ship.

Answer to Problem 54E

The x -coordinate of the position of the ship is $110.279$ .

Explanation of Solution

Given information:

The time difference between the pulses from the transmitting stations is 1000 microseconds. The ship is travelling on a hyperbolic path with coordinates $\left(x,75\right)$ .

Calculation:

Consider the given diagram.

  

Let consider a hyperbola. In conic, the foci is the distance between two points along the major axis and it is represented by $ae$ . The length of the semi-major axis is $a$ and the eccentricity is $e$ . Let c is the distance between station 1 and station 2.

From the diagram, calculate the value of the semi-major axis and major axis.

  $\begin{array}{c}2c=300\\ c=150\end{array}$

Calculate the semi-major axis.

  $\begin{array}{c}2a=0.001\times 186000\\ 2a=186\\ a=93\end{array}$

Calculate the length of the minor axis.

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={150}^2-{93}^2\\ =13851\end{array}$

Write the expression for hyperbolic path.

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Substitute the values in the above equation.

  $\begin{array}{c}\frac{x^2}{{\left(93\right)}^2}-\frac{{75}^2}{13851}=1\\ x\approx 110.279\end{array}$

Therefore, the x -coordinate of the position of the ship is $110.279$ .

(b)

To determine

To find: The distance between the ship and station 1.

Answer to Problem 54E

The distance between the ship and station 1 is 57 miles.

Explanation of Solution

Given information:

The time difference between the pulses from the transmitting stations is 1000 microseconds. The ship is travelling on a hyperbolic path with coordinates $\left(x,75\right)$ .

Calculation:

Consider the given diagram.

  

When the ship reaches to shore, the distance between the ship and station 1 is the distance from the focus to the center.

  $\text{Distance}=a-c$

Substitute the values in the above expression.

  $\begin{array}{c}\text{Distance}=150-93\\ =57\text{miles}\end{array}$

Therefore, the distance between the ship and station 1 is 57 miles.

(c)

To determine

To find: The time difference between the pulses.

Answer to Problem 54E

The time difference between the pulses is $0.001$ seconds.

Explanation of Solution

Given information:

The bay is at a distance of 30 miles from station 1.

Calculation:

Consider the given diagram.

  

Since, the bay is at a distance of 30 miles from station 1. The coordinate of the bay will be $\left(120,0\right)$ .

Calculate the distance between the bay and station 2.

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-150-120\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(-270\right)}^2}\\ =270\text{miles}\end{array}$

Since the pulses travels at a speed of 186000 miles per second. The time difference between the pulses will be equal to,

  $\begin{array}{c}\frac{270}{186000}-\frac{30}{186000}=\frac{240}{186000}\\ =\frac{1}{775}\\ \approx 0.001\text{seconds}\end{array}$

Therefore, the time difference between the pulses is $0.001$ seconds.

(d)

To determine

To find: The position of the ship if the ship is at a distance of 60 miles offshore.

Answer to Problem 54E

The position of the ship is $\left(144.2,60\right)$ .

Explanation of Solution

Given information:

The ship is 60 miles offshore.

Calculation:

Consider the given diagram.

  

In part (c), the time difference between the pulses is $0.001$ seconds. Since, the ship is 60 miles offshore. So, the y -coordinate is 60. Calculate the semi-major axis.

  $\begin{array}{c}2a=\left(\frac{1}{775}\right)186000\\ 2a=240\\ a=120\end{array}$

Calculate the length of the minor axis.

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={150}^2-{120}^2\\ =8100\end{array}$

Write the expression for the hyperbolics path.

  $\begin{array}{c}\frac{x^2}{{120}^2}-\frac{{60}^2}{8100}=1\\ x\approx 144.2\end{array}$

Therefore, the position of the ship is $\left(144.2,60\right)$ .