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Physics ships direction

markosheehan

Member
Jun 6, 2016
136
a cruiser is 12 km due west of a destroyer . the cruiser is travelling in a direction 60 degrees N of E at 30 km hr^-1 and the destroyer is travelling at 15 km hr^-1. find the direction the destroyer must travel in order to get as close as possible to the cruiser


there is complicating way to do this however there is also a easy way and i am trying to find out the easy way to do it.
the answer is 120 degrees .
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,491
a cruiser is 12 km due west of a destroyer . the cruiser is travelling in a direction 60 degrees N of E at 30 km hr^-1 and the destroyer is travelling at 15 km hr^-1. find the direction the destroyer must travel in order to get as close as possible to the cruiser


there is complicating way to do this however there is also a easy way and i am trying to find out the easy way to do it.
the answer is 120 degrees .
Let's see... it usually helps to draw a picture...
\begin{tikzpicture}[scale=0.5,>=stealth]
\def\x{110};
\def\t{0.3};
\node (C) at (0,0) {};
\node (D) at (12,0) {};
\node (C1) at ({30*\t*cos(60)}, {30*\t*sin(60)}) {};
\node (D1) at ({12+15*\t*cos(\x)}, {15*\t*sin(\x)}) {};
\fill (C) circle (0.1) node[below] {C};
\fill (D) circle (0.1) node[below] {D};
\fill (C1) circle (0.1) node[above] {C'};
\fill (D1) circle (0.1) node[above right] {D'};
\draw (D) -- +(3,0);
\draw[->,thick] (C) +(1,0) node[above right] {$60^\circ$} arc (0:60:1);
\draw[->,thick] (D) +(1,0) node[above right] {$x$} arc (0:\x:1);
\draw (C) -- node[below] {12} (D);
\draw[->,ultra thick,blue] (C) -- node[above left] {30t} (C1);
\draw[->,ultra thick,blue] (D) -- node[above right] {15t} (D1);
\draw[->,ultra thick,red] (D1) -- node[above right] {d} (C1);
\end{tikzpicture}

The standard way to solve this, is to write $d^2$ as a function of both $x$ and $t$, and then find the critical point.
That is, set both partial derivatives to zero.
$$d^2=(30t\cos(60^\circ)-(12 + 15t\cos(x)))^2 + (30t\sin(60^\circ)-15t\sin(x))^2 \\ \pd{}t(d^2)=0 \\ \pd{}x(d^2)=0
$$
That's indeed a bit of work.
Wolfram says that the solution is $t=0.4$, $x=120^\circ$, and $d=6$.

\begin{tikzpicture}[scale=0.5,>=stealth]
\def\x{120};
\def\t{0.4};
\node (C) at (0,0) {};
\node (D) at (12,0) {};
\node (C1) at ({30*\t*cos(60)}, {30*\t*sin(60)}) {};
\node (D1) at ({12+15*\t*cos(\x)}, {15*\t*sin(\x)}) {};
\fill (C) circle (0.1) node[below] {C};
\fill (D) circle (0.1) node[below] {D};
\fill (C1) circle (0.1) node[above] {C'};
\fill (D1) circle (0.1) node[above right] {D'};
\draw (D) -- +(3,0);
\draw[->,thick] (C) +(1,0) node[above right] {$60^\circ$} arc (0:60:1);
\draw[->,thick] (D) +(1,0) node[above right] {$x$} arc (0:\x:1);
\draw (C) -- node[below] {12} (D);
\draw[->,ultra thick,blue] (C) -- node[above left] {30t} (C1);
\draw[->,ultra thick,blue] (D) -- node[above right] {15t} (D1);
\draw[->,ultra thick,red] (D1) -- node[above right] {d} (C1);
\end{tikzpicture}

Now here's an interesting observation.
At the minimum the vector $\vec {DD'}$ is parallel to the distance vector $\vec d$.
That makes sense, since as long as there is an angle between them, we can improve the distance by reducing that angle by changing $x$ appropriately.
It means we can eliminate either $x$ or $t$ from the expression for the distance, and then take its derivative and set it to zero.
Still some work though.