The Tragedy of the SS Brilliant

Let's ignore the drama and intrigue and get down to the heart of the problem:

1. The SS Brilliant (blue) is moving in a rectilinear path with uniform velocity $v$ .
2. The pirate ship (red) is initially located at a distance $D$ from the SS Brilliant, with a velocity of $2v$ pointing towards the SS Brilliant .

Initial Positions and Velocities

1. As the SS Brilliant moves forward, the pirate ship maintains its speed but changes its direction of motion to point towards the SS Brilliant at all times.
2. Eventually, the pirate ship reaches the SS Brilliant by following some curved trajectory.

Trajectories

Now, how do we find the distance $D$ ? That is the crux of the entire problem. How about we try using the basic definitions of kinematics to get started?

Velocities at time t - Equation 1

From this, we see that at time $t$ , the SS Brilliant has moved a distance $vt$ from the coastline. Meanwhile, the pirate ship has a velocity $2v\sin \theta$ in the positive y direction. This velocity is the rate of change of position along the y-axis. Hence, we can write:

$\dfrac{dy}{dt}=2v\sin \theta$

$dy=2v \sin \theta dt$

At the moment when the two ships collide, their positions along the y-direction will be the same. Since we know that this position is $vt$ for the SS Brilliant , we know it is the same for the pirate ship. Hence, we set the limits of integration from $0$ to $vt$ :

$\int_{0}^{vt} dy = 2v\int_{0}^{t} \sin \theta dt$

$vt=2v\int_{0}^{t} \sin \theta dt$

$\Longrightarrow \int_{0}^{t} \sin \theta dt =\dfrac{t}{2}$

But this doesn't help us get $D$ , right? $D$ does not appear anywhere here! That's why we need another equation, this time, by resolving the velocities of the ships along the line connecting them:

Velocities at time t - Equation 2

Now, let $R$ be the distance between the ships at any time $t$ . The rate of change of this distance will be only due to the components of velocities along the line R, i.e. only the $v \sin \theta$ and $2v$ velocities can possibly alter the value of $R$ . Hence, we can write:

$\dfrac{dR}{dt}=v\sin \theta -2v$

$dR=(v\sin \theta -2v)dt$

The initial distance between the ships is $D$ while the final distance is $0$ since they collide. Hence, we set the limits of integration from $D$ to $0$ :

$\int_{D}^{0} dR = \int_{0}^{t} (v\sin \theta -2v)dt$

$\int_{D}^{0} dR = -2vt+v\int_{0}^{t} \sin \theta dt$

But we know from our first equation that $\int_{0}^{t} \sin \theta dt=\dfrac{t}{2}$ :

$\int_{D}^{0} dR = -2vt+\dfrac{vt}{2}$

$\Longrightarrow D=2vt-\dfrac{vt}{2}$

We know the velocity is 8 knots, and that the time taken for them to meet is 3 hours (from 08:00 to 11:00). Hence,

$\boxed{D=36}$

In three hours, the SS Brilliant moved $3 \times 8=24$ nautical miles in the positive y direction.

Pythagorean theorem

Applying the Pythagorean theorem, the distance the pirate ship has to travel back is $\sqrt{24^2+36^2}=12\sqrt{13}$ .

At a speed of 12 knots, this distance will take $\sqrt{13}$ hours to travel, which is approximately 3 hours and 36 minutes.

At 11:00, the pirates boarded. At 14:14 (3 hours and 14 minutes later), they headed back to their cove and 3 hours and 36 minutes later, they arrived at the cove.

$\Longrightarrow 17:50$