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{\bf Question}

The ship HMS Alpha is steaming due north with a speed
10kmh$^{-1}$, whilst HMS Beta is steaming north east with the same
speed.  Find the velocity of HMS Alpha relative to HMS Beta.  If
at noon HMS is positioned 10km due west of HMS Beta show that
subsequently the position of HMS Alpha relative to HMS Beta is
$\ds 10 \left[-\left(1 + \frac{t}{\sqrt 2}\right) {\bf i} +
\left(1 - \frac{1}{\sqrt 2}\right)t{\bf j}\right]$, where ${\bf i}
$ and ${\bf j}$ are the unit vectors oriented east and north and
$t$ is the time elapsed past noon.  At what time was HMS Alpha due
south of HMS Beta?

\vspace{.25in}

{\bf Answer}


Situation at noon:

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\put(0.9,0.75){$\alpha$}

\put(2.9,0.7){$\beta$}

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$\ds {\bf v}_{\alpha \beta} = {\bf v}_{\alpha} - {\bf v}_{\beta} =
10{\bf j} - \frac{10}{\sqrt 2}({\bf i} + {\bf j}) = 10 \left(-
\frac{1}{\sqrt2}{\bf i} + \left(1 - \frac{1}{\sqrt 2} \right) {\bf
j}\right)$

Position of $\alpha$ relative to $\beta$ satisfies $\ds
\frac{d}{dt} {\bf r}_{\alpha \beta} = {\bf v}_{\alpha \beta}$

Thus $\ds {\bf r}_{\alpha \beta} = {\bf v}_{\alpha \beta}t + {\bf
a}$ where $\bf a$ is a constant of integration.

At t = 0, ${\bf r}_{\alpha \beta} = -10{\bf i}$ (position of
$\alpha$ relative to $\beta$ at noon)

\begin{eqnarray*} {\bf r}_{\alpha \beta} & = & \frac{10}{\sqrt 2}
[-{\bf i} + (\sqrt 2 - 1) {\bf j}]t - 10 {\bf i} \\ & = &
\frac{10}{\sqrt 2} [-{\bf i}(t + \sqrt 2) + t(\sqrt 2 - 1) {\bf
j}] \end{eqnarray*}

$\alpha$ is due south of $\beta$ when ${\bf r}_{\alpha \beta}$ has
no $\bf i $ component

i.e. at t = -$\sqrt 2 \mathrm{hrs} \approx -1.414\mathrm{hrs}$ or
at 10.35 am

The distance is $\ds \frac{10}{\sqrt 2} \sqrt 2(\sqrt 2 - 1) =
414$km



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