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{
\bf Dr. Z's Math251 Handout \#12.2 [Vectors]
}

By Doron Zeilberger

{\bf Problem Type 12.2a}: A person walks due west (or east)
on the deck of a ship at $a$ mi/h. This ship is moving
north (or south, or whatever) at $b$ mi/h. Find the speed
and direction of the person relative to the surface of the water.

{\bf Example Problem 12.2a}: 
A girl walks due east
on the deck of a ship at $5$ mi/h. This ship is moving
south at $12$ mi/h. Find the speed
and direction of the girl relative to the surface of the water.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the velocity vectors of the person
and the ship. Add them up to get the composite velocity.

(East is in the direction of the {\bf positive} $x$ axis so
it is a postive multiple of the vector {\bf i} 
(or $\langle 1,0 \rangle$) 
West is in the direction of the {\bf negative} $x$ axis so
it is a postive multiple of the vector -{\bf i} 
(or $\langle -1,0 \rangle $) 
North is in the direction of the {\bf positive} $y$ axis so
it is a postive multiple of the vector {\bf j} (or $\langle 0,1 \rangle$) 
South is in the direction of the {\bf negative} $y$ axis so
it is a postive multiple of the vector -{\bf j} (or $\langle 0,-1 \rangle $)).


&
{\bf 1.} 
The velocity vector of the girl is $5 \bf{i}$,
and the velocity vector of the ship is $-12 \bf{j}$.
So the velocity vector relative to the water is the sum
$5 \bf{i}-12 \bf{j}$ (or $\langle 5,-12 \rangle$).
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{\bf 2.} If the composite velocity vector is
$a {\bf i} + b {\bf j}$ (or $\langle a,b \rangle$), then
the {\bf speed} is the {\bf magnitude},
which is $\sqrt{a^2+b^2}$,
and the direction is at angle $\theta=\tan^{-1} b/a$
with the positive $x$-axis (alias East).
&
{\bf 2.} 
speed= $\vert 5 \bf{i}-12 \bf{j} \vert=
\sqrt{5^2+12^2}=13$
and the direction is $\theta=\tan^{-1} (-12/5)=-\tan^{-1} 12/5$.

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