Question 199619
 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 = {{{15/4}}}
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