Question 166382: 84.)Gone Fishing. Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4 - mph current, it took her 20 minutes longer to get there than to return. how fast will her boat go in still water?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=Debbie's rate in still water
Time to travel upstream=5/(r-4) (Note: against the current, we subtract speed of the current; with the current, we add.)
Time to travel downstream=5/(r+4)
Now we are told that it took 20 min (or 0.33 hr) longer to go upstream than downstream,so our equation to solve is:
(5/(r-4))-0.33=5/(r+4) multiply each term by (r-4)(r+4)
5(r+4)-0.33(r-4)(r+4)=5(r-4) get rid of parens (distributive law)
5r+20-0.33r^2+5.28=5r-20 subtract 5r from and add 20 to both sides
5r-5r+20+20-0.33r^2+5.28=5r-5r-20+20 collect like terms
-0.33r^2+45.28=0 divide each term by -0.33
r^2-137.21=0 add 137.21 to each side
r^2=137.21 take sqrt of each side
r= 11.7 mph--------------Debbies speed in still water
Disregard the negative value for r
CK
(5/(11.7-4))-0.33=5/(11.7+4)
(5/7.7)-0.33=5/15.7
0.319~~~0.319
Hope this helps---ptaylor
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