An aircraft took off point A(xºN, 15ºE) at 0720h, local time. It flew due West to another point B(xºN, 75ºW) a distance of 5005 km from A. After a stopover ot 1 hour 30 mimutes at point B. the aircraft took off and flew for 3 hours 40 minutes due South to a point C. The aircraft maintained an average speed of 910 km/h for the journey fom A to B and also from B to C (Take π = 22 and the radius of the earth to be 6370 km) 7

Question

An aircraft took off point A(xºN, 15ºE) at 0720h, local time. It flew due West to another point B(xºN, 75ºW) a distance of 5005 km from A.
After a stopover of 1 hour 30 minutes at point B, the aircraft took off and flew for 3 hours 40 minutes due South to a point C. The aircraft maintained an average speed of 910 km/h for the journey from A to B and also from B to C
(Take π = ²²/₇and the radius of the earth to be 6370 km)

1. Calculate the:
   • position of point B. 
   • position of point C 
2. Determine the local time at point C when the aircraft arrived.            

Answer 1

1.  
   1. θ= 75 + 15 = 90
       90   x 2 x 22 x 6,370 cos a = 5005
      360            7
      cos a= 0.500
      a= 60
      B(60 N, 75W)
   2. speed x time= (^θ/₃₆₀) x 2πR
      910 x 3hrs 40 mins = (^θ/₃₆₀)2 x ²²/₇ x 6370
      θ= 30
      new latitude = 60N − 30 = 30N
      C(30 N, 75 W)
2. local time C when departing from A(0720hr)
   1⁰   = 4 min
   90⁰ =?
   ⁹⁰/₁ x 4 min = 6hrs
   time in c when departing from A
   = 0720 hrs − 6 hrs
   = 0120 hrs
   arrival time = departure time + travelling time
   = 0120hrs + 33mins + 1 1/2 + 3hrs 40mins
   = 0703 hrs or 7. 03 am