SOLUTION: I had an algebra test this morning and I am still trying to figure out this problem. I'm sure it is not near as hard as I am making it. The problem read: Bob leaves the camp at 3:0

Question 199619: I had an algebra test this morning and I am still trying to figure out this problem. I'm sure it is not near as hard as I am making it. The problem read: Bob leaves the camp at 3:00 paddling downstream at 4mph (stillwater) with a 2mph current. Joe leaves the camp at 4:30 paddling downstream twice as fast as Bob (the water is flowing at the same rate). What time will Bob and Joe meet?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Bob leaves the camp at 3:00 paddling downstream at 4mph (stillwater) with a 2mph current.
Joe leaves the camp at 4:30 paddling downstream twice as fast as Bob (the water
is flowing at the same rate).
What time will Bob and Joe meet?
:
Let t = the travel time of Bob
then
(t-1.5) = travel time of Joe (leaves 1.5 hrs after Bob)
;
Bob's speed with the current: 4 + 2 = 6 mph
Joe's speed with the current: 8 + 2 = 10 mph
:
When they meet they will have traveled the same distance. Write a dist equation:
Dist = speed * time
:
B's dist = J's dist
6t = 10(t-1.5)
6t = 10t - 15
15 = 10t - 6t
4t = 15
t =
t = 3.75 hrs from 3:00 is 6:45 when they meet
:
:
Check solution by ensuring they did, in fact, travel the same distance.
6*3.75 = 22.5 mi
10*2.25 = 22.5 mi
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