# 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

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 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(15649)   (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