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it took a faster runner 10 sec. longer to run a distance of 1500 ft. than it took a slower runner to run a distance of 1000ft. if the rare of the faster runner was 5ft/sec more than the role of slower runner, what was the rate of each. |
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Answer» It took a faster runner 10 seconds longer to run a distance of 1500 feet than it took a slower runner to run a distance of 1000 feet.Let t = time in sec for the faster runner to run 1500 ftthen(t-10) = time of the slower runner to run 1000 ft:if the rate of the faster runner was 5 ft/s more than the rate of the slower runner, what was the rate of each?":Write a rate equation: Rate = dist/time::FAST rate - slow rate = 5 ft/sec1500%2Ft - 1000%2F%28t-10%29 = 5:Multiply equation by t(t-10), results1500(t-10) - 1000t = 5t(t-10):1500t - 15000 - 1000t = 5t^2 - 50t:500t - 15000 = 5t^2 - 50t:0 = 5t^2 - 50t - 500t + 15000A QUADRATIC equation5t^2 - 550t + 15000 = 0Simplify, divide by 5t^2 - 110t + 3000 = 0Factor to(t - 50)(t - 60) = 0two solutionst = 50 sect = 60 sec:ASSUME t = 501500/50 = 30 ft/sec for the fast guyand1000/40 = 25 ft/sec for the slow guy (5 ft/sec slower) |
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