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Rashid [163]
3 years ago
5

Problem page a long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. one hour later, a

cyclist starts at the beginning of the trail and travels at a rate of 14 miles per hour. what is the amount of time that the cyclist travels before overtaking the runner? do not do any rounding.
Mathematics
1 answer:
alukav5142 [94]3 years ago
5 0
The thing you have to figure out about this is the distance for each person and the time it takes for the biker to meet the runner.  The rates we are told.  The formula is distance = rate times time.  Let's do that for the runner first.  His rate is 6 so the formula so far is d = 6t.  Now let's work on the time.  If the biker left an hour later than the runner, then the runner has been running an hour more than the biker.  Therefore, the runner's time is t + 1.  Hold off on the distance part til we do for the biker what we just did for the runner.  The biker's rate is 14, and we already decided that his time is t.  His equation is d = 14t.  Now at the exact moment the biker meets the runner their distances are the same.  So if the equation for the runner is d = 6t + 6 and the equation for the biker is d = 14t and their distances are the same, by the transitive property, their rates and times are the same as well, meaning we set them equal to each other and solve for t.  6t + 6 = 14t.  6 = 8t and t = 3/4.  This means that it took 45 minutes for the biker to meet the runner.
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