SOLUTION: Suppose that a cyclist began a 513 mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state. If

Question 1027739: Suppose that a cyclist began a 513 mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 9.5 hours and the car traveled 38.2 mph faster than the bicycle, find the average rate of each
Found 2 solutions by ankor@dixie-net.com, mananth:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Suppose that a cyclist began a 513 mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state.
If the bicycle and car met after 9.5 hours and the car traveled 38.2 mph faster than the bicycle, find the average rate of each
:
let s = the speed of the bicycle
then
(s+38.2) = the speed of the car
:
When they meet the sum of their distances will be 513 mi
Write a distance equation; dist = time * speed
:
bike dist + car dist = 513 mi
9.5s + 9.5(s+38.2) = 513
9.5s + 9.5s + 362.9 = 513
19s = 513 - 362.9
s = 150.1/19
s = 7.9 mph is the speed of the bike
then
7.9 + 38.2 = 46.1 mph is the speed of the car
:
:
Check this, find the sum of the actual distances each traveled
9.5 *7.9 = 75.05 mi
9.5*46.1 = 437.95 mi
-------------------------
total dist: 513.0 mi
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
cyclist x mph
car x+ 38.2 mph
..
They are moving towards each other

So add up their speed.

combined speed = x+ 1 x 38.2
( 2 x + 38.2 )
Time = 9.5 hours
Distance = 513 miles
Distance = speed * time
( 2 x 38.2 )* 9.5 = 513
19.00 x + 362.9 = 513
19 x = 513 + -362.9
19 x = 150.1
/ 19
x= 7.9 mph
cyclist = 7.9 mph
car 7.9 + 38.2 = 46.1 mph
m.ananth@hotmail.ca

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