183_notes:examples:relativemotion

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183_notes:examples:relativemotion [2014/07/11 12:47] – [Solution] caballero183_notes:examples:relativemotion [2014/11/16 08:05] (current) pwirving
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 === Lacking === === 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 $v_{p/a} = 255 dfrac{m}{s}$).+  * 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 === === Approximations & Assumptions ===
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   * 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.   * 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.    * 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 ==== ==== Solution ====
  
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 $${v}_{p/a,x}^2 + {v}_{p/a,y}^2 = 1$$ $${v}_{p/a,x}^2 + {v}_{p/a,y}^2 = 1$$
 +
 +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.
 +
 + 
  • 183_notes/examples/relativemotion.1405082835.txt.gz
  • Last modified: 2014/07/11 12:47
  • by caballero