Problema Solution
two air planes leave at the same airport at the same time. The airplane flying east travels at an average speed that is 20 mph faster than the airplane flying north. If the two airplanes are 680 mi apart after 1 hr, find the average speed of each airplane to the nearest mile per hour.
Answer provided by our tutors
let
v1 = the speed of the plane traveling east, v1>0
v2 = the speed of the plane traveling north, v2>0
t = 1 hr
the airplane flying east travels at an average speed that is 20 mph faster than the airplane flying north
v1 = 20 + v2
since speed = distance/time => distance = time*speed and using the Pythagorean Theorem we have
(v1*t)^2 + (v2*t)^2 = 680^2
plug v1 = 20 + v2 into the last equation
(20 + v2)^2 + v2^2 = 680^2
by solving we find
v2 = 471 mph
click here to see the step by step solution of the equation
v1 = 471 + 20
v1 = 491 mph
the average speed of the planes are 471 mph and 491 mph.