SOLUTION: a tortoise crawling at the rate of.1 mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at the rate of 5 mi/h. How may feet must the
Question 334396: a tortoise crawling at the rate of.1 mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at the rate of 5 mi/h. How may feet must the hare run to catch the tortoise? Answer by jrfrunner(365) (Show Source):
You can put this solution on YOUR website! the tortoise travels at 0.1 mph therefore after 1/2hour (distance=rate*time)
the tortoise will have travelled =(0.1mph)*(0.5 hour)=0.05 miles*(5280ft/mile)=264 feet.
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the hare will have to make up 264 feet plus the additional distance the tortoise makes after the hare starts running.
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tortoise distance: =0.1*0.5+0.1*t
Hare distance: 5*t
find t when both these distances are equal
5*t=0.05+0.1*t
t=.05/4.9=0.010204 hour
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So distance hare has to make up: d=rt= 5*0.010204=0.051002 miles*(5280ft/mile)=269.4 ft the hare has to make up.
