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The one about flying to Reno You plan to fly a small airplane from San Jose Airp

ID: 1652391 • Letter: T

Question

The one about flying to Reno You plan to fly a small airplane from San Jose Airport to Reno-Tahoe International Airport. Using a map you find that on the map San Jose Airport if 475 miles north and 12.2 miles west of the origin, and Reno-Tahoe airport is at 623 miles north and 132 miles east of the maps origin. The National Weather Service reports that there will be steady winds of 15 knots coming from the west. Your airplane cruises with an airspeed of 120 mph, and you can approximate its entire journey to be with this airspeed (i.e. neglect acceleration during take-off and landing). What compass heading (i.e. what direction with North = 0 degree, East = 90 degree, etc.) should you fly? How long will the flight be? Given the limited precision of the numbers used in your calculation (i.e. the significant digits) how close would you expect to be to the airport if you just pointed your plane in the calculated direction and waited the calculated time?

Explanation / Answer

Given, San hose airport
475 miles north, 12.2 miles east of origin
Reno Tahoe airport
623 miles north, 132 miles east of origin

Wind speed, w = 15 knots , coming from west

Airspeed, v = 120 mph

1. let i be a unit vector along east, j be a unit vector along north
   so vector from Airport A to B is
   r = 119.8 i + 148 j
   now, w = 15 i knots = 17.2617 i mph
   speed of plane wrt ground = speed of plane wrt air + speed of air wrt ground
   so, v' = v + 15 i
   now let v = 120(cos(theta)i + sin(theta)j)
   then v' = (120cos(theta) + 17.2617)i + 120sin(theta)j
   also, r = v'*t

   119.8 i + 148 j = [(120cos(theta) + 17.2617)i + 120sin(theta)j] * t
   comparing we get

   120sin(theta)*t = 148
   (120cos(theta) + 17.2617)t = 119.8

   (120cos(theta) + 17.2617)/120*sin(theta) = 0.809459
   120cos(theta) + 17.2617 = 97.135*sin(theta)
   14400*cos^2(theta) + 297.966 + 43499.484*cos(theta) = 9435.208225*(1 - cos^2(theta))
   23835.208225*cos^2(theta) + 43499.484*cos(theta) -9137.242225 = 0
   solving for cos theta
   cos(theta) = 0.19022
   theta = 79.03 deg

   so the plane must fly 79.03 deg north of east
2. 120sin(theta)*t = 148
   t = 1.2562
   so the flight will be 1.2562 hours long

3. the distance accuracy of reaching the plane, dr
   dr/r = d(theta)/theta
   so as the accuracy of every measurement is upto 0.1 miles, we can expect to reach the destination by an accuracy near to it