SOLUTION: 4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still wa

Algebra ->  Rational-functions -> SOLUTION: 4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still wa      Log On


   

Question 199564: 4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still water?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = speed of boat in still water



d=rt Start with the distance-rate-time formula


40=%28x-3%29t Plug in d=40 and r=x-5. This equation represents the upstream journey.


40%2F%28x-5%29=t Divide both sides by x-5 to isolate "t";


So the expression for the time it takes to go upstream can be represented by the expression 40%2F%28x-5%29


-----------------------------------


d=rt Go back to the distance-rate-time formula

40=%28x%2B5%29t Plug in d=40 and r=x%2B5. This equation represents the downstream journey


40%2F%28x%2B5%29=t Divide both sides by x%2B5 to isolate "t";


So the expression for the time it takes to go downstream can be represented by the expression 40%2F%28x%2B5%29


Now simply add the two time expressions to get: 40%2F%28x-5%29%2B40%2F%28x%2B5%29


40%2F%28x-5%29%2B40%2F%28x%2B5%29=6 Now set that expression equal to the total time of 6 hours


40%28x%2B5%29%2B40%28x-5%29=6%28x%2B5%29%28x-5%29 Multiply every term by the LCD %28x%2B5%29%28x-5%29 to clear the denominators.


40%28x%2B5%29%2B40%28x-5%29=6%28x%5E2-25%29 FOIL


40x%2B200%2B40x-200=6x%5E2-150 Distribute


40x%2B200%2B40x-200-6x%5E2%2B150=0 Subtract 6x%5E2 from both sides. Add 150 to both sides.


-6x%5E2%2B80x%2B150=0 Combine like terms


Notice we have a quadratic in the form of Ax%5E2%2BBx%2BC where A=-6, B=80, and C=150


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%2880%29+%2B-+sqrt%28+%2880%29%5E2-4%28-6%29%28150%29+%29%29%2F%282%28-6%29%29 Plug in A=-6, B=80, and C=150


x+=+%28-80+%2B-+sqrt%28+6400-4%28-6%29%28150%29+%29%29%2F%282%28-6%29%29 Square 80 to get 6400.


x+=+%28-80+%2B-+sqrt%28+6400--3600+%29%29%2F%282%28-6%29%29 Multiply 4%28-6%29%28150%29 to get -3600


x+=+%28-80+%2B-+sqrt%28+6400%2B3600+%29%29%2F%282%28-6%29%29 Rewrite sqrt%286400--3600%29 as sqrt%286400%2B3600%29


x+=+%28-80+%2B-+sqrt%28+10000+%29%29%2F%282%28-6%29%29 Add 6400 to 3600 to get 10000


x+=+%28-80+%2B-+sqrt%28+10000+%29%29%2F%28-12%29 Multiply 2 and -6 to get -12.


x+=+%28-80+%2B-+100%29%2F%28-12%29 Take the square root of 10000 to get 100.


x+=+%28-80+%2B+100%29%2F%28-12%29 or x+=+%28-80+-+100%29%2F%28-12%29 Break up the expression.


x+=+%2820%29%2F%28-12%29 or x+=++%28-180%29%2F%28-12%29 Combine like terms.


x+=+-5%2F3 or x+=+15 Simplify.


So the possible solutions are x+=+-5%2F3 or x+=+15

However, you can't have a negative speed. So the only answer is x+=+15


=================================================================
Answer:


So the speed of the boat in still water is 15 mph.