Question

a tortoise crawling at a rate of 0.1 mph, the hare wants to rest 30 more minutes before chasing the turtle at 5 mph. how many feet must the hare run to catch the turtle?

Answer 1

during the 30 minutes, the tortoise travels

1/2 (.1) = .05 mi

so, if the hare runs for t hours,

5t = .05 + .1t
t = .0102 hours
at 5 mph, the hare runs 5*.0102 = .051 miles = 269.28 feet

check: the turtle has moved an additional .1*.0102 = 0.00102 miles, or .051 miles in all

The two distances are the same.