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Hard algebra question

Posted on 2014-02-23
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Last Modified: 2014-02-24
Elias and Emma are travelling up the San Marcos River in separate canoes. Elias, averaging 3 mph, had gone 5 miles when Emma first put her canoe into the water.
   a) How fast must Emma paddle her canoe to catch up to Elias by the end of the 15 mile long canoe route?
      
   b) How long will it take Emma to reach Elias?
      
   c) On the way up the river, the current slowed Elias’ canoe to 3 mph. But on the way back he found that he could paddle 6.5 mph. How fast was the current?

   d) How fast could Elias have paddled his canoe if there were no current?
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Question by:JohnMac328
6 Comments
 
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Expert Comment

by:ozo
ID: 39880996
15 miles = t * Emma
15 miles = 5 miles + t * (3 miles / hour)
Emma = 15 miles / ((15 miles - 5 miles)/(3 miles / hour))

3 mph = Elias - current
6.5 mph = Elias + current
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Author Comment

by:JohnMac328
ID: 39881015
I am having trouble with part C, what equation can I use to solve this part?
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Assisted Solution

by:Don Thomson
Don Thomson earned 250 total points
ID: 39881191
Current = C
current Up = -C
Current Down = +C

Speed Down = 3
Speed Up = 6.5

Speed Diff =3.5
Current must be 3.5/2

3+C = 6.5 -C

C=6.5-3 -C

2C=3.5
C=1.75

Nocurrent   = 3 +1.75 or 6.5-1.75

You do the rest
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Accepted Solution

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Fred Marshall earned 250 total points
ID: 39881237
a) Elias' speed = 3mph
    Elias time at 5 miles = 5/3 = 1.67 hours. (not important here)
    Distance remaining for Elias 15-5=10 miles
    Time for Elias to complete 10 mi / 3 mph = 3.33 hours.
    Time for Emma to complete is the same 3.33 hours.
    Emma's speed to complete in 3.33 hours is 15 mi /3.33 hrs = 4.5 mph

b) See the above in (A) 3.33 hours.

c) 3mph = speed over water - current
   6.5mph=speed over water + current
   subtracting gets 3.5mph = 2*current so current = 3.5/2=1.75mpg

d) speed over ground 3mph
    speed over water = 3mph + current = 3 + 1.75 = 4.75 mph
     or                           = 6.5mph - current = 6 - 1.75 = 4.75 mph
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Expert Comment

by:Don Thomson
ID: 39882297
I think you mean mph  not mpg  ;)
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Author Closing Comment

by:JohnMac328
ID: 39882693
Thanks!
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