Two ships A and B originally at a distance d from each other depart at the same time from a straight coastline. Ship A moves along a straight line perpendicular to the shore while ship B constantly heads for ship A, having at each moment the same speed as the latter. After a sufficiently great interval of time the second ship will obviously follow the first one at a certain distance. Find the distance.

To solve the problem of the two ships, we need to analyze their movements and the relationship between them over time. Ship A moves straight out to sea, while Ship B is always adjusting its course to head directly towards Ship A. This creates a fascinating dynamic that can be modeled mathematically.

Understanding the Movement of the Ships

Let's break down the situation:

• Ship A: It moves straight away from the coastline at a constant speed, say \( v \).
• Ship B: It also moves at speed \( v \) but constantly adjusts its direction to head towards Ship A.

Initially, the distance between the two ships is \( d \). As time progresses, Ship A moves further away from the shore, while Ship B tries to close the gap by moving towards Ship A.

Analyzing the Trajectories

To visualize this, imagine Ship A moving vertically away from the coastline, while Ship B is always aiming for Ship A. The path of Ship B will not be a straight line but rather a curved trajectory that continuously adjusts as Ship A moves further away.

At any given moment, the angle between the direction of Ship B's movement and the line connecting the two ships changes. This means that Ship B is not only moving towards Ship A but also has to account for the distance Ship A is gaining.

Mathematical Formulation

We can use calculus to find the distance between the two ships after a certain time \( t \). Let’s denote the distance from the coastline to Ship A as \( y(t) = vt \), where \( t \) is the time elapsed. The distance between the two ships can be represented as a right triangle, where:

• The vertical leg is \( y(t) \) (the distance Ship A has traveled).
• The horizontal leg is the distance Ship B has traveled towards Ship A.

Using the Pythagorean theorem, the distance \( D \) between the two ships can be expressed as:

D(t) = sqrt(d^2 + (vt)^2)

Finding the Distance After a Long Time

As time goes on, Ship A continues to move away from the coastline, and Ship B will eventually reach a point where it can no longer close the gap completely. The distance \( D \) will stabilize at a certain value. To find this distance, we can analyze the limit as \( t \) approaches infinity.

As \( t \) becomes very large, the term \( (vt)^2 \) will dominate the equation, leading us to:

D(t) ≈ vt

However, we need to consider the angle at which Ship B is moving. The effective distance that Ship B can maintain from Ship A will be a constant value, which can be derived from the geometry of the situation. The final distance \( d' \) between the two ships can be shown to be:

d' = d / sqrt(2)

Conclusion

In summary, after a sufficiently long time, Ship B will maintain a distance of \( d / \sqrt{2} \) from Ship A. This fascinating problem illustrates the interplay between constant speed and changing direction, leading to a stable distance despite the continuous movement of both ships. This concept can be applied to various real-world scenarios, such as pursuit problems in navigation and robotics.