Question 1056047: Janet set off on a bicycle trip at 630am at an average rate of 26 miles per hour. her husband . Bob, promised to bring her some lunch.
he set off by car 2 hours after janet left driving at an average speed of 40 miles per hour. At what time did bob catch up with Janet?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39627)   (Show Source):
You can put this solution on YOUR website! Time  begins at 6:30AM, but time t will be some other, positive value and this quantity can be easier to use than a time point on a line.
Bob's catchup distance is d.
NOW,  and you will solve for it.
SPEED TIME DISTANCE
JANET 26 t+2 d
BOB 40 t d
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website! Janet set off on a bicycle trip at 630am at an average rate of 26 miles per hour. her husband . Bob, promised to bring her some lunch.
he set off by car 2 hours after janet left driving at an average speed of 40 miles per hour. At what time did bob catch up with Janet?
Let time it takes Bob to get to catchup point, be T
Then time it takes Janet to get to catchup point is: T + 2
We then get the following DISTANCE equation: 40T = 26(T + 2)
40T = 26T + 52
40T – 26T = 52
14T = 52
T, or time it takes Bob to get to catch-up point =  , or 3.714286 hours, or 3 hours, 42.85714286 minutes, or approximately 3 hours, 43 minutes
Time Bob caught Janet: 
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