2. (6 points) Consider problem 3.71: 3.71 » An airplane pilot sets a compass cou

Question

2. (6 points) Consider problem 3.71: 3.71 » An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for 0.500 h, shoe finds herself over a town 120 km west and 20 km south of her starting position. Helpful information: “ sets a compass course due west , is yet another way to say "...the airplane points due west...." 2.a. (2 pts) Write the general equation that relates the 3 velocity vectors to each other. 2.b. (4 pts) Fill-in the blank cells of this table, taking N as +y and E as +f as usual: Name Magnitude Direction x-component y-component vP/E v P/A v A/E don't do this onedon't do this one unknown unknown unknown unknown Note: you are not solving for anything. You are only doing these important setup steps.

Explanation / Answer

airspeed of the plane, v = 220 km/h ( speed wrt air)
let speed of wind wrt ground be u

1. speed of plane wrt ground = v'
   then , v' = v + u

2. let i and j be unit vectors along east and north respectively
   then , after t = 0.5 hr
   displacement, r = v'*t
   but v' = v + u
   and v = - 220 i
   let u = (ai + b j)
   then form the given data
   -120i - 20j = (-220i + ai + bj)*0.5
   comparing coefficients
   -110 + 0.5a = -120
   a = -20
   and b = -40
   so u = -20i -40 j

   also, v' = vP/E = (-120i - 20j)/0.5 = -240i - 40j

   so vP/E x component = -240 km/hr
   vP/E y component = -40 km/hr

   vP/A
   Magnitude = 220 km/h
   directionj = -i ( west)
   x-component = -220 km/h
   y component = 0

   v A/E
   x-component = -20 km/h
   y - component = -40 km/h
   magnitude = sqroot(20^2 + 40^2) = 44.72 kmn/h
   directikon = arctan(40/20) = 63.43 deg south of west

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