I have been reading the book "A Course in Classical Physics 1 - Mechanics" by Alessandro Bettini. I have reached a problem in the first chapter that I am having trouble with.
The problem statement goes :
We are on a ship travelling at 10 kn heading east. We see another ship, which we know moves at 20 kn to North, 6 miles distant in the South direction. What is the minimum distance the two ships will be (without changing their courses)? After how much time ? On the sea distances are measured in nautical miles and velocities in knots (1 kn = 1 mile/h). Assume for the mile the round figure of 1800 m.
The answer key for the problem states :
She will cross at 3 miles towards stern in 18′
This is not the answer that I am getting and wanted some help to see what I could be doing wrong. My solution so far is below.
$\require{amsmath}$ $\require{cases}$ $\DeclareMathOperator*{\argmin}{argmin}$
We can draw the scenario :
Here ship A is my ship and ship B is the other ship. We work in a frame attached to ship A with the positive y axis pointing North and the positive x axis pointing East. Let $\vec r_{A}(t)$ be the position of ship A as a function of time $t$ and $\vec r_{B}(t)$ be the position of ship B. Let $\vec v_{A}(t)$ and $\vec v_{B}(t)$ be the velocities of the two ships. The velocity functions have units of knots (miles per hour) and the position functions have units of miles. $t$ has units of hours.
We have : \begin{align} \vec v_{A}(t) & = (0,0) \\\\ \vec v_{B}(t) & = (-10,20) \end{align} So : \begin{align} \vec r_{A}(t) & = (0,0)\\\\ \vec r_{B}(t) & = (-10t,20t-6) \end{align} and the displacement vector between the two ships is : \begin{equation} \vec r_{AB}(t) = \vec r_{B}(t) - \vec r_{A}(t) = (-10t,20t - 6) = \vec r_{B}(t) \end{equation} So the distance between the two ships is given by : \begin{align} d_{AB}(t) = |\vec r_{AB}(t)| & = \sqrt{(-10t)^{2} + (20t-6)^{2}}\\\\ & = \sqrt{100t^{2} + (20t - 6)^{2}} \end{align} We see : \begin{align} (20t - 6)^{2} & = (20t - 6)(20t - 6)\\\\ & = 400t^{2} - 120t - 120t + 36 \\\\ & = 400t^{2} - 240t + 36 \end{align} So : \begin{align} d_{AB}(t) & = \sqrt{100t^{2} + 400t^{2} - 240t + 36}\\\\ & = \sqrt{500t^{2} - 240t + 36}\\\\ & = \sqrt{f(t)} \end{align} Where : \begin{equation} f(t) = 500t^{2} - 240t + 36 \end{equation} It seems : \begin{equation} \hat t = \argmin_{t \geq 0} d_{AB}(t) \Leftrightarrow \hat t = \argmin_{t \geq 0} f(t) \end{equation} as the square root function is increasing. We see : \begin{equation} f'(t) = \frac{d}{dt}\left( 500t^{2} - 240t + 36 \right) = 1000t - 240 \end{equation} and : \begin{align} f'(t) = 0 & \Leftrightarrow 1000t - 240 = 0 \\\\ & \Leftrightarrow 1000t = 240 \\\\ & \Leftrightarrow t = \frac{240}{1000} \\\\ & \Leftrightarrow t = \frac{24}{100} \\\\ & \Leftrightarrow t = \frac{6}{25} \end{align} So : \begin{align} f'(t) > 0 \text{ when } t > \frac{6}{25} \\\\ f'(t) \leq 0 \text{ when } t \leq \frac{6}{25} \end{align} So $f(t)$ is monotone decreasing before $t = \frac{6}{25}$ and increasing afterwards. This means the minimum of $f(t)$ is at $t = \frac{6}{25}$. So : \begin{equation} \argmin_{t \geq 0} d_{AB}(t) = \frac{6}{25} \text{ hours } \end{equation} And we see : \begin{align} \frac{6}{25} \text{ hours } & = \left[ \frac{6}{25} \cdot 60 \right] \text{ minutes }\\\\ & = \frac{360}{25} \text{ minutes }\\\\ & = 14.4 \text{ minutes } \end{align} Which is not equal to the correct answer of 18 minutes.
Can someone show me where I am going wrong ?
