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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.