Problema Solution
At noon, two airplanes leave an airport and travel in opposite directions at the same altitude. The slower plane is flying at 360 mph, and the faster plane is traveling at 450 mph. When will the two planes be 1350 miles apart?
Answer provided by our tutors
let
v1 = 360 mph the average speed of the slower plane
v2 = 450 mph the average speed of the faster plane
d = 1350 mi the distance between the planes
t = the time of the travel
v1*t + v2*t = d
360t + 450t = 1350
by solving we find:
t = 5/3 hr
t = 5*60/3
t = 100 min
t = 1 hr 40 min
click here to see the step by step solution of the equation:
1 hr 40 min after noon is 1:40 pm.
At 1:40 pm the two planes will be 1350 miles apart.