Question

Two airplanes are spotted on a radar map located at the coordinates (72,- 5) and (3./2,5/5). Assume the planes are flying in exactly the same direction. The plane behind is flying at 1050 mph and at an altitude of 2500 ft., and the other plane is flying at 850 mph and at an altitude of 3500 ft. As the air traffic controller, you are given the task to determine how many miles it will take for the plane behind to be directly below the plane flying at 3500 11.

Answer 1

The distance between two points (21, yı) and (22, 92) is V(x1 - x2)2 + (y1 - y2)2. Thus, the initial horizontal distance between the two planes = V(V2 – 312)2+(-V5 – 5V5)2 = V(-272)2 +(–6V/5)2 = V8 + 180 = V188 miles

Now, let the required number of miles be x. Let us call the plane flying behind at an altitude of 2500 ft the first plane. Let us call the other plane the second plane. Then, time taken for the first plane to travel x+V188 miles = time taken for the second plane to travel x miles x + 188 distance distance · speed =- = time =- 1050 time speed 850x + 850/188 = 1050x = 850188 = 1050x – 850x 850188 = 200. BX= 850V188 = 58.273 miles 200 It will take 58.273 miles for the first plane to be directly below the second plane.