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--- 183_notes:examples:relativemotion [2014/07/11 02:29] – created caballero
+++ 183_notes:examples:relativemotion [2014/11/16 08:05] (current) – pwirving
@@ Line 10: / Line 10: @@
   * The pilot intends to fly due west.
-  * The plane experiences a crosswind with a speed of 10.0 $\dfrac{m}{s}$, which is directed due south.
+  * The plane experiences a crosswind with a speed of $|v_{a/g}| = 10.0 \dfrac{m}{s}$, which is directed due south.
 === Lacking ===
-  * The top speed of a Boeing 747 is unknown, but can be [[http://lmgtfy.com/?q=top+speed+of+747|found online]] (920  $\dfrac{km}{h}$  or 255  $dfrac{m}{s}$ ).
+  * The top speed of a Boeing 747 is unknown, but can be [[http://lmgtfy.com/?q=top+speed+of+747|found online]] (920  $\dfrac{km}{h}$  or $v_{p/a} = 255 \dfrac{m}{s}$).
 === Approximations & Assumptions ===
@@ Line 26: / Line 26: @@
   * The velocities of the plane relative to the air, the air relative to the ground, and the plane relative to the ground are represented in the following diagram.
-<WRAP todo>Add vector addition diagram</WRAP>
+{{ 183_notes:planerelativemotion.png?350 }}
   * The relative velocity equation for three objects is:  $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$  where  $\vec{v}_{A/C}$  is the velocity of object A with respect to object C, etc.
 ==== Solution ====
+The problem can be described vectorially using the relative velocity equation:
+ $\vec{v}_{p/g} = \vec{v}_{p/a} + \vec{v}_{a/g}$
+The pilot requires that the velocity of the plane with respect to the ground be directed due west. If you measure positive  $\theta$  counterclockwise with respect to the east, the plane's velocity relative to the ground should only have a negative  $x$ -component. So the equation above becomes (in 2D),
+ $\langle -|v_{p/g}|,0 \rangle = |v_{p/a}|\hat{v}_{p/a} + \langle 0, -|v_{a/g}|\rangle$
+We can break this vector equation into two scalar equations:
+ $-|v_{p/g}|=|v_{p/a}|{v}_{p/a,x}$
+ $0=|v_{p/a}|{v}_{p/a,y}-|v_{a/g}|$
+where  ${v}_{p/a,x}$  and  ${v}_{p/a,y}$  are the components of the unit vector in the direction of the velocity of the plane with respect to the air. Thus, they satisfy this equation.
+ ${v}_{p/a,x}^2 + {v}_{p/a,y}^2 = 1$
+You can rewrite the above equation by using what the unit vector components are equal to (in the previous two equations),
+ $\left(-\dfrac{|v_{p/g}|}{|v_{p/a}|}\right)^2+\left(\dfrac{|v_{a/g}|}{|v_{p/a}|}\right)^2= 1$
+ ${|v_{p/g}|}^2+{|v_{a/g}|}^2= {|v_{p/a}|}^2$
+So, the speed that the plane has with respect to the ground is slower than its air speed, which agrees with the representation above.
+ ${|v_{p/g}|}= \sqrt({|v_{p/a}|}^2-{|v_{a/g}|}^2) = \sqrt{(255 \dfrac{m}{s})^2 - (10 \dfrac{m}{s})^2} = 225 \dfrac{m}{s}$
+The angle the compass should read can be determined from the above representation. The tangent of the angle (as measured from the negative  $x$ -axis is given by,
+ $\tan \theta = \dfrac{|v_{a/g}|}{|v_{p/g}|}$
+Hence,
+ $\theta = \tan^{-1} \left(\dfrac{|v_{a/g}|}{|v_{p/g}|}\right) = \tan^{-1} \left(\dfrac{10 \dfrac{m}{s}}{225 \dfrac{m}{s}}\right) = 2.5^{\circ}$
+which is 2.5 $^{\circ}$  north of west or 177.5 $^{\circ}$  from east measured counterclockwise.