SOLUTION: Need equation:Jane leaves school and walks home at a rate of 4 mi/hr. 15 minutes later, Bill finds a book that Jane forgot and runs at a rate of 8 mi/hr. How many minutes will it t

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Question 1004968: Need equation:Jane leaves school and walks home at a rate of 4 mi/hr. 15 minutes later, Bill finds a book that Jane forgot and runs at a rate of 8 mi/hr. How many minutes will it take Bill to catch up with Jane?
I know she has walked 1 mile in 15 minutes (1/4 hour @ 4 mi/hr is 1 mile). At his pace, it will take 1/2 that time to walk the same distance. But do I consider that she is walking away from him at same time? Don't know how to put in equation format.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Jane's head start in miles:
+4%2A%281%2F4%29+=+1+ mi
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Now imagoine that you start a stopwatch
when Bill leaves.
Let +t+ = time in hours on stopwatch
when Bill catches Jane
Let +d+ = distance in miles Bill travels
until he catches up with Jane
-----------------------------------
Bills equation:
(1) +d+=+8t+
Jane's equation:
(2) +d+-+1+=+4t+
----------------
Substitute (1) into (2)
(2) +8t+-+1+=+4t+
(2) +4t+=+1+
(2) +t+=+1%2F4+
It will take Bill 15 min to catch Jane
-----------------------------
check:
(1) +d+=+8t+
(1) +d+=+8%2A%281%2F4%29+
(1) +d+=+2+
and
(2) +d+-+1+=+4t+
(2) +d+-+1+=+4%2A%28+1%2F4%29+
(2) +d+-+1+=+1+
(2) +d+=+2+
OK