two airplanes depart simultaneously from an airport.One flies due south, the other flies due east at a rate 20 mi/h faster than that of the other airplane.

Question

two airplanes depart simultaneously from an airport.One flies due south, the other flies due east at a rate 20 mi/h faster than that of the other airplane. After one hour , radar indicates that the airplanes are approximately 450 mi apart. What is the ground speed of each airplane?

Answer 1

v = speed of slow plane

v + 20 = speed of fast plane

t = 1 hr

 

You have a right triangle with a base of (v + 20)t, a height of vt, and a hypotenuse of 450.

 

Then

 

(v + 20)²t² + v²t² = 450²

 

t = 1 ⇒ (v + 20)² + v² = 450²

 

v² + 40v + 400 + v² = 202500

 

2v² + 40v - 202100 = 0

 

v² + 20v - 101050 = 0

 

v = (1/2){-20 ± √[20² - 4(1)(-101050)]}

 

v = (1/2)(-20 ± 636.08)

 

Use the positive square root, since we want  a positive speed (magnitude of velocity).

 

v = 308.04 mph (south)

v + 20 = 328.04 mph (east)