SOLUTION: 2 planes travel toward eachother from cities that are 1615km apart. the 2 planes pass eachother in the air after 3hrs and 48min, the slower plane was traveling at a rate of speed t

Question 168610: 2 planes travel toward eachother from cities that are 1615km apart. the 2 planes pass eachother in the air after 3hrs and 48min, the slower plane was traveling at a rate of speed that was 80kh more than 1/2 the rate of speed of the faster plane assuming they left at the same time at what rate of speed was each plane traveling?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
2 planes travel toward each other from cities that are 1615km apart. the 2
planes pass each other in the air after 3hrs and 48min, the slower plane was
traveling at a rate of speed that was 80kh more than 1/2 the rate of speed
of the faster plane assuming they left at the same time, at what rate of
speed was each plane traveling?
:
Let s = rate of the faster plane
then
(.5s+80) = rate of the slower plane
:
Change 3 hr 48 min to hrs; 3 + 48/60 = 3.8 hrs
:
When they meet. they will have traveled a total of 1615 km
They both have a travel time of 3.8 hr
:
Write a distance equation: dist = time * speed
:
Fast speed dist + slow speed dist = 1615 km
3.8s + 3.8(.5s+80) = 1615
:
3.8s + 1.9s + 304 = 1615
:
5.7s = 1615 - 304
:
5.7s = 1311
s =
s = 230 km/hr is the fast plane
then
.5(230) + 80 = 195 km/hr is the slow one
:
:
Check solution by finding the total distance
3.8*230 = 874
3.8*195 = 741
---------------
total d = 1615

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