Homework Help: Speed and distance

Posted by Marysvoice on Saturday, February 9, 2008 at 10:58am.

I can't figure this one out

During a trip, a canoe travles 47 miles at r speed. The second part of the trip the canoe travels 9 miles, but 5 mph slower. The total trip is 5 hrs. What is the speed of each part of the trip?

I start with 47/r+9/r-5=5hrs
Then I multiply both sides by the LCD?
r(r-5)(47/r+9/r-5)=5
47(r-5)+9r=5r(r-5)
47r-???+9r=3r^2? I am lost here
Maybe-
47x5r=235?
47r-235+9r=5r-4r?
Quadratic-
5r^2-60r+235=0???

Speed and distance - Quidditch, Saturday, February 9, 2008 at 12:08pm
You have it figured out.
47(r-5)+9r=5r(r-5)
47r-235+9r=5r^2-25r
Rearranging and combining you had an error.
I got 5r^2-81r+235=0
Solve using the quadratic formula. One of the solutions of the quadratic will be less than 5 which would imply that the canoe would change directions for the (r-5) rate.
- Speed and distance - domonic, Saturday, February 9, 2008 at 6:12pm
  hi quidditch!!! remember me??
- Speed and distance - domonic, Saturday, February 9, 2008 at 6:13pm
  hi quidditch!!! remember me??
Speed and distance - drbob222, Saturday, February 9, 2008 at 12:11pm
start with 47/r+9/r-5=5hrs OK to here. I will rewrite to see it more clearly.
[47/r] + [9/(r-5)] = 5
multiply both sides by r(r-5) to get
47(r-5) + 9r = 5(r)(r-5)
47r - 235 + 9r = 5r^2 - 25r
47r - 235 + 9r -5r^2 + 25r = 0
-5r^2 + 81r -235 = 0
I changed the sign here but it isn't necessary.
5r^2 -81r + 235 = 0
and solve by the quadratic formula. I et r = 12.4 and if I remember the problem that is mph. Then r-5 = 7.4 mph.
I hope this helps.

Then I multiply both sides by the LCD?
r(r-5)(47/r+9/r-5)=5
47(r-5)+9r=5r(r-5)
47r-???+9r=3r^2? I am lost here
Maybe-
47x5r=235?
47r-235+9r=5r-4r?
Quadratic-
5r^2-60r+235=0???
Speed and distance - Marysvoice, Saturday, February 9, 2008 at 1:18pm
I GOT IT!!!!! THANK YOU!!! THANK YOU!!! THANK YOU!!!

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