SOLUTION: In a 1-mile race, the winner crosses the finish line 12 ft ahead of the second-place runner and 22 ft ahead of the third-place runner. Assuming that each runner maintains a consta
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Question 138275This question is from textbook Blitzer College Algebra
: In a 1-mile race, the winner crosses the finish line 12 ft ahead of the second-place runner and 22 ft ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner? (5280 ft in a mile) This question is from textbook Blitzer College Algebra
You can put this solution on YOUR website! In a 1-mile race, the winner crosses the finish line 12 ft ahead of the second-place runner and 22 ft ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner? (5280 ft in a mile)
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How far did the the 2nd and 3rd place runners travel when the winner crossed the the finish line (5280 ft)?
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5280 - 12 = 5268 ft, traveled by the 2nd place runner
5280 - 22 = 5258 ft, traveled by the 3rd place runner
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Use a ratio equation =
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Cross multiply:
5268x = 5280 * 5258
x =
x = 5269.977 traveled by the 3rd place runner when the 2nd place runner crosses the finish line
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5280 - 5269.977 = 10.023 ft is the distance the 2nd place runner beats the 3rd place runner
:
This makes sense because they were 10 ft apart when the winner crossed the line
3rd place runner is a little over 10 ft behind the 2nd place guy, when he crosses the finish line because, he is slightly slower
