- Determine the constant speed \(v\) by calculating the rate of change between the given time and distance data points, resulting in \(v = 21.25\) nautical miles per hour.
- Calculate the initial distance \(d_0\) at \(t = 0\) by using the linear model \(d(t) = vt + d_0\) with a known data point, yielding \(d_0 = 10.5\) nautical miles.
- Find the distance at \(11\) hours by evaluating the linear function at \(t = 11\), resulting in \(d(11) = 244.25\) nautical miles.
- Final results: Distance at 11 hours is \(\boxed{244.25}\), initial distance is \(\boxed{10.5}\), and speed is \(\boxed{21.25}\).
Explanation
- Analysis of the Data and Modeling the Motion
The problem provides a table of time-distance data points for a cruise ship moving at a constant speed relative to a lighthouse. We have the data points: \((2, 53)\), \((4, 95.5)\), \((6, 138)\), \((10, 223)\), and \((12, 265.5)\). Given that the speed is uniform, the distance \(d\) as a function of time \(t\) follows a linear relationship of the form \(d(t) = vt + d_0\), where \(v\) represents the uniform speed and \(d_0\) is the initial distance from the lighthouse at \(t = 0\). We need to determine the speed \(v\), the initial distance \(d_0\), and the distance at \(t = 11\) hours.
- Calculating the Speed of the Ship
Since the ship travels at a constant speed, the speed \(v\) is equivalent to the slope of the linear equation \(d(t) = vt + d_0\). We can calculate the slope using any two points from the table, for instance, \((2, 53)\) and \((4, 95.5)\):
\[v = \frac{d_2 - d_1}{t_2 - t_1} = \frac{95.5 - 53}{4 - 2} = \frac{42.5}{2} = 21.25 \text{ nautical miles per hour}.\]
Verifying this with points \((6, 138)\) and \((10, 223)\) gives:
\[v = \frac{223 - 138}{10 - 6} = \frac{85}{4} = 21.25 \text{ nautical miles per hour}.\]
The cruise ship maintains a constant speed of \(21.25\) nautical miles per hour.
- Determining the Initial Distance
With the speed \(v = 21.25\) known, we use one of the data points, such as \((2, 53)\), to find the initial distance \(d_0\). Substituting into the linear equation \(d(t) = vt + d_0\):
\[53 = 21.25(2) + d_0\]
\[53 = 42.5 + d_0\]
\[d_0 = 53 - 42.5 = 10.5 \text{ nautical miles}.\]
Thus, the cruise ship was \(10.5\) nautical miles from the lighthouse at the start of its journey.
- Calculating the Distance at 11 Hours
Now that we have the complete linear equation \(d(t) = 21.25t + 10.5\), we can find the distance at \(t = 11\) hours:
\[d(11) = 21.25(11) + 10.5\]
\[d(11) = 233.75 + 10.5\]
\[d(11) = 244.25 \text{ nautical miles}.\]
After \(11\) hours, the cruise ship is \(244.25\) nautical miles from the lighthouse.
Examples
This problem illustrates how linear relationships are used in navigation to predict the position of a moving vehicle based on its speed. By tracking a ship's distance from a fixed point over time, navigators can determine its current path and future location. This mathematical modeling is fundamental to fields such as logistics, flight tracking, and transportation management, where maintaining a schedule and understanding movement patterns are essential for safe and efficient operations. For example, if a cargo ship is moving at a steady rate, port authorities can calculate exactly when it will arrive at the terminal by using the linear equation derived from its initial position and speed.