Question

The captain of a man-of-war saw, one dark night, a privateersman crossing his path at right angles and at a distance ahead of \(c\) miles. The privateersman was making \(a\) miles an hour, while the man-of-war could make only \(b\) miles in the same time. The captain's only hope was to cross the track of the privateersman at as short a distance as possible under his stern, and to disable him by one or two well-directed shots; so the ship's lights were put out and her course altered in accordance with this plan. Show that the man-of- war crossed the privateersman's track \(\frac{c}{b} \sqrt{a^{2}-b^{2}}\) miles astern of the latter. If \(a=b\), this result is absurd. Explain.

Short answer

Question: Calculate the distance astern (y) at which the man-of-war crosses the privateersman's track, given their initial distance (c) and their speeds (a and b). Answer: The distance astern (y) at which the man-of-war crosses the privateersman's track can be calculated using the formula: \[y = \frac{c}{b} \sqrt{a^{2}-b^{2}}\] Keep in mind that if both ships have the same speed (a = b), the result will be absurd and the man-of-war will not be able to cross the privateersman's track and disable the privateersman.

Step-by-step solution

Calculate the distance between two ships

Let's call the point where the privateersman is at the moment of the observation A, the point where the man-of-war is at the same moment B, and the point where the man-of-war crosses the privateersman's track C. Using the Pythagorean theorem, we can calculate the distance AC, which we will call x. Consider right-angled triangle ABC, with AC=x, BC=c, and AB=d. Using the Pythagorean theorem: \[d^{2} = x^{2} + c^{2}\]

Calculate the times taken by both ships to reach point C

Since the speeds of the ships are given, we can find the times taken by both ships to reach point C. For the privateersman ship travel from A to C, we have: \[ t_{1} = \frac{x}{a}\] For the man-of-war ship travel from B to C: \[t_{2} = \frac{d}{b}\]

Relate the times and the distance

Since both ships travel along the same path and reach point C at the same time, we have: \[t_{1} = t_{2}\] Therefore, \[\frac{x}{a} = \frac{d}{b}\] Now, we can write d in terms of x: \[d = \frac{b}{a}x\]

Substitute d in the Pythagorean equation

Now we will substitute d from step 3 into the equation from step 1: \[\left(\frac{b}{a}x\right)^{2} = x^{2} + c^{2}\] Solve for x: \[x^{2} \left(\frac{b^{2}}{a^{2}}-1\right) = -c^{2}\] \[x = c\sqrt{\frac{a^{2}-b^{2}}{b^{2}}}\]

Calculate the distance astern

The distance astern is the distance that the man-of-war crosses the privateersman's track. Therefore, the distance astern, y, is: \[y = \frac{c}{b} \sqrt{a^{2}-b^{2}}\]

Explain the absurdity when a = b

If a = b, it means both ships have the same speed, and the man-of-war will never be able to cross the privateersman's track to get a chance to shoot. In this case, the equation for y becomes: \[y = \frac{c}{b} \sqrt{a^{2}-a^{2}} = \frac{c}{b} \cdot 0 = 0\] This result does not make sense because it implies that the man-of-war is crossing the privateersman's track right at his stern, which is only possible if their initial positions were already crossing, which is not the case in the problem statement. The absurdity arises because the man-of-war cannot disable the privateersman in this case, making the result invalid.

Key concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects. In our exercise, kinematics is used to determine how the two ships, the man-of-war and the privateersman, move towards point C. The fundamental kinematic parameter is speed, which describes how fast an object moves. Here, the privateersman travels at a speed of \(a\) miles per hour, while the man-of-war moves at \(b\) miles per hour. The objective is for the man-of-war to cross the path of the privateersman at the closest point astern.
The kinematic approach involves calculating the time it takes for both ships to reach point C. This means:

• For the privateersman: This time is \(t_1 = \frac{x}{a}\), where \(x\) is the distance from A to C.
• For the man-of-war: This time is \(t_2 = \frac{d}{b}\), where \(d\) is the initial distance from B to C.

The condition \(t_1 = t_2\) ensures both ships arrive at C simultaneously, showing the intersection of their paths through kinematics.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, especially involving right-angled triangles. It relates the sides of the triangle with the equation \(c^2 = a^2 + b^2\). In this problem, we represent the situation using a triangle.
The vertices of the triangle are

• A: The position of the privateersman ship when first spotted.
• B: The position of the man-of-war ship when the privateersman is spotted.
• C: The point where the man-of-war crosses the privateersman's path.

Using the theorem, we formulate the relationship between the sides as \(d^2 = x^2 + c^2\), where
- \(d\) is the hypotenuse running from B to C
- \(x\) is one leg from A to C
- \(c\) is the other leg from B to C.
Substituting \(d = \frac{b}{a}x\) derived from our kinematics analysis allows us to express \(x\) in terms of known quantities, bridging kinematics and geometry through this theorem.

Relative Motion

Relative motion considers the motion of objects as observed from another moving object. It helps in understanding how one object moves concerning another. In the problem, relative motion is crucial because the man-of-war's goal is to cross behind the privateersman.
Even though both ships travel directly without stopping, their movement paths are angled to each other, impacting how they perceive each other's motion. When calculating the closest distance astern, we are actually considering the relative paths involved:

• The privateersman moves straight along its intended path.
• The man-of-war adjusts its course to strategically pass under the privateersman's stern, considering the relative speeds.

The relative velocity approach helps to identify the point where one ship overtakes or crosses the path of another — allowing us to calculate the specific farthest trailing distance \(y = \frac{c}{b} \sqrt{a^2 - b^2}\). This formula accounts for the adjusted positions due to their respective speeds, ensuring the man-of-war crosses as intended.