SOLUTION: 1. A ship leaves a harbor sailing at 28 mph. A plane leaves 6-1/2 hours later. At what rate must it fly to overtake the ship in an hour and 15 min.? 2. A man flew to another ci

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: 1. A ship leaves a harbor sailing at 28 mph. A plane leaves 6-1/2 hours later. At what rate must it fly to overtake the ship in an hour and 15 min.? 2. A man flew to another ci      Log On

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Question 1060571: 1. A ship leaves a harbor sailing at 28 mph. A plane leaves 6-1/2 hours later. At what rate must it fly to overtake the ship in an hour and 15 min.?
2. A man flew to another city for a meeting at a rate of 260 mph. He returned by train at a rate of
60 mph. If his total travel time was 4 hours, what was his flying time?
I am a home school mom trying to help my daughter with these 2 problems. I would appreciate any help. Thank you very much! Tara
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
These problems are in the category of
" related rate " problems. For word problems,
it's very important to identify their category.
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(1)
What is the head start in miles for the ship?
+d%5B1%5D+=+28%2A6.5+
+d%5B1%5D+=+182+ mi
------------------------
Let +d+ = the distance in miles the plane
travels until it catches up with the ship
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Let +s+ = the flying speed of the plane so
that the plane catches up with the ship in
+1.25+ hrs
-----------------------------------------
Equation for the plane:
(1) +d+=+s%2A1.25+
Equation for the ship:
(2) +d+-+182+=+28%2A1.25+
-----------------------------
Substitute (1) into (2)
(2) +1.25s+-+182+=+28%2A1.25+
(2) +1.25s+=+35+%2B+182+
(2) +1.25s+=+217+
(2) +s+=+173.6+
The plane's flying rate is 173.6 mi/hr
-------------------------------------
check the answer:
(1) +d+=+s%2A1.25+
(1) +d+=+173.6%2A1.25+
(1) +d+=+217+
and
(2) +d+-+182+=+28%2A1.25+
(2) +217+-+182+=+28%2A1.25+
(2) +35+=+35+
OK
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---------------------------------------
(2)
Let +t+ = his flying time in hrs
+4+-+t+ = his time traveling by train
Let +d+ = the distance to the city
------------------------------------
Equation for traveling by train:
(1) +d+=+60%2A%28+4+-+t+%29+
Equation for flying:
(2) +d+=+260t+
-------------------------
substitute (2) into (1)
(1) +260t+=+60%2A%28+4+-+t+%29+
(1) +260t+=+240+-+60t+
(1) +320t+=+240+
(1) +t+=+.75+ hrs
His flying time is 3/4 of an hour, or 45 minutes
--------------------------
check the answer
(2) +d+=+260t+
(2) +d+=+260%2A.75+
(2) +d+=+195+ mi
and
(1) +d+=+60%2A%28+4+-+t+%29+
(1) +d+=+60%2A%28+4+-+.75+%29+
(1) +d+=+60%2A3.25+
(1) +d+=+195+ mi
OK
Hope this helps