SOLUTION: John left school at 2:30. He travels 3 miles per hour. His brother left school 5 minutes later. He travels 3.6 miles per hour. At what time will Johns's brother catch up with h

Question 137245: John left school at 2:30. He travels 3 miles per hour. His brother left school 5 minutes later. He travels 3.6 miles per hour. At what time will Johns's brother catch up with him?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
John left school at 2:30. He travels 3 miles per hour. His brother left school 5 minutes later. He travels 3.6 miles per hour. At what time will John's brother catch up with him?
:
Change 5 min to hrs: 5/60 = 1/12 hr
:
Let t = John's travel time (in hrs), when his brother catches up
then
(t-) = his brother's travel time
:
We know when this is accomplished, they will have traveled the same distance.
Write a distance equation: Dist = speed * time
:
Brothers dist = John's dist
3.6(t-) = 3t
:
3.6t - .3 = 3t; (3.6 * 1/12 = .3)
:
3.6t - 3t = .3
:
.6t = .3
t =
t = hr
Brother catches up at 3 o'clock
:
Check solution by finding the distance of both (Brother's time: 25/60 = .4167 hr)
Brother: 3.6 * .4167 = 1.5 mi
John: 3 * .5 = 1.5 mi

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