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--- 183_notes:examples:relativemotion [2014/07/11 12:57] – [Solution] caballero
+++ 183_notes:examples:relativemotion [2014/11/16 08:05] (current) – pwirving
@@ 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 ====
@@ Line 55: / Line 54: @@
 So, the speed that the plane has with respect to the ground is slower than its air speed, which agrees with the representation above.
-$${|v_{p/g}|}= \sqrt({|v_{p/a}|}^2-{|v_{a/g}|}^2) = \sqrt{(225 \dfrac{m}{s})^2 - (10 \dfrac{m}{s})^2} = 225 \dfrac{m}{s}$$
+$${|v_{p/g}|}= \sqrt({|v_{p/a}|}^2-{|v_{a/g}|}^2) = \sqrt{(255 \dfrac{m}{s})^2 - (10 \dfrac{m}{s})^2} = 225 \dfrac{m}{s}$$
+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.