\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

A river of width $d$ has uniform current with speed $u$.  If the
ferry can make speed $v$ relative to the water at what angle
should the ferry set off from one bank in order to arrive directly
opposite on the far bank?

\medskip

How long does it take the ferry to cross the river?

\bigskip

{\bf Answer}

$ {} $

\setlength{\unitlength}{.5in}
\begin{picture}(10,5)
\put(1.5,1.5){\line(1,0){5}}

\put(1.5,4.5){\line(1,0){5}}

\put(8,3){\vector(1,0){.5}}

\put(8,3){\vector(0,1){.5}}

\put(8.6,3){\makebox(0,0){\bf i}}

\put(8,3.8){\makebox(0,0){\bf j}}

\put(1,1.5){\vector(0,1){3}}

\put(1,2.5){\vector(0,-1){1}}

\put(0.75,3){\makebox(0,0){$d$}}

\put(4,1){\makebox(0,0){Bank}}

\put(4,5){\makebox(0,0){Bank}}

\put(3.5,2.5){\circle*{.075}}

\put(3.5,2.5){\line(0,1){1}}

\put(3.5,2.5){\vector(0,1){0.5}}

\put(3.5,2.5){\line(1,1){1}}

\put(3.5,2.5){\vector(1,1){0.5}}

\put(3.5,2.2){\makebox(0,0){Ferry}}

\put(3.2,3){\makebox(0,0){{\bf v}$_{FB}$}}

\put(4.4,3){\makebox(0,0){{\bf v}$_{FR}$}}

\put(2,3){\makebox(0,0){River}}

\put(6.5,3){\line(-1,0){1}}

\put(6.5,3){\vector(-1,0){0.5}}

\put(6.7,3){\makebox(0,0){\bf u}}

\end{picture}


Use the equation $\ds {\bf v}_{FB} = {\bf v}_{FR} + {\bf v}_R$;

where $\begin{array}{cc}  {\bf v}_{FB} & {\rm is\ the\ velocity\
of\ the\ ferry\ relative\ to\ the\ bank} \\  {\bf v}_{FR} &
\rm{is\ the\ velocity\ of\ the\ ferry\ relative\ to\ the\ river}
\\  {\bf v}_R & {\rm is\ the\ velocity\ of\ the\ river\ relative\ to\
the\ bank} \end{array}$


\begin{eqnarray*} {\bf v}_{FB} = {\bf v}_{F}{\bf j};
\hspace{.2in}& & {\bf v}_R = -u{\bf i};  \hspace{.2in} {\bf
v}_{FR} = v({\bf i}\cos \theta + {\bf j} \sin \theta) \\
\Rightarrow {\bf v}_{FB} & = & v({\bf i}\cos \theta + {\bf j} \sin
\theta) - u{\bf i} \\ \Rightarrow 0 & = & {\bf i}(v \cos \theta -
u) + {\bf j}(v \sin \theta - v_{FB}) \\ \Rightarrow \theta & = &
\cos ^{-1} \frac{u}{v} \\ {\rm and \ \ } v_{FB} & = & v \sin
\theta = v \sqrt{1 - \frac{u^2}{v^2}} = \sqrt{ v^2 - u^2}
\end{eqnarray*} Time to cross the river is $\ds \frac{d}{v_{FB}} =
\frac{d}{\sqrt{ v^2 - u^2}}$



\end{document}
