SOLUTION: Please solve this following problem. A tortoise that can walk three miles in two hours starts walking at 12:00. At 3:00, a hare that can walk fifteen miles in four hours sets ou

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Please solve this following problem. A tortoise that can walk three miles in two hours starts walking at 12:00. At 3:00, a hare that can walk fifteen miles in four hours sets ou      Log On


   



Question 1054014: Please solve this following problem.
A tortoise that can walk three miles in two hours starts walking at 12:00. At 3:00, a hare that can walk fifteen miles in four hours sets out on the trail of the tortoise. At what time will the hare catch up to the tortoise and how far are they from their starting point when they meet?
Time =
distance=

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
What is the tortoise's head start in miles?
Note that 12 to 3 PM is +3+ hrs
+d%5B1%5D+=+%28+3%2F2+%29%2A3+
+d%5B1%5D+=+9%2F2+ mi
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Start a stop watch when the hare leaves.
Let +t+ = time on the stop watch in hrs
Let +d+ = distance in miles the hare travels
until it catches up with the tortoise.
Stop the stop watch when hare catches up
with tortoise
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The tortoise's equation:
(1) +d+-+9%2F2+=+%283%2F2%29%2At+
The hare's equation:
(2) +d+=+%28+15%2F4+%29%2At+
-----------------------
Substitute (2) into (1)
(1) +%28+15%2F4+%29%2At+-+9%2F2+=+%283%2F2%29%2At+
Multiply both sides by +4+
(1) +15t+-+18+=+6t+
(1) +9t+=+18+
(1) +t+=+2+ hrs
The hare catches up with the tortoise
at 3 PM plus 2 hrs = 5 PM
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(2) +d+=+%28+15%2F4+%29%2At+
(2) +d+=+%28+15%2F4+%29%2A2+
(2) +d+=+15%2F2+
They are 7.5 miles from their starting
point when they meet
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check answer:
(1) +d+-+9%2F2+=+%283%2F2%29%2At+
(1) +15%2F2+-+9%2F2+=+%283%2F2%29%2A2+
(1) +6%2F2+=+6%2F2+
OK