Question 651130

"A tortoise crawling at a rate of 0.1mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at a rate of 5mi/h. How many feet must the hare run to catch the tortoise?"


The time taken by the tortoise to get to the "catch-up" point, equals the time it takes the hare to get to the "catch-up point," PLUS 30 minutes ({{{1/2}}} hour, or 0.5 hour). This is written as:


D/0.1 = D/5 + 0.5
5D = .1D + .25 ------ Multiplying by LCD, 0.5
5D - .1D = .25
4.9D = .25
D, or distance hare needs to travel to catch up to the tortoise = {{{.25/4.9}}}, or 0.5102 mile


Distance for hare to cover = 0.05102 mile
5,280 ft = 1 mile


Therefore, we have: {{{5280/F = 1/0.05102}}}, with F being distance in feet the hare needs to run

Cross-multiplying, we get: F = 5,280 * 0.05102


F, or distance the hare needs to run to catch up to the tortoise = {{{highlight_green(269.39)}}} feet


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