SOLUTION: During the first part of a trip a canoeist travels 48 miles at a certain speed. The canoeist travels 19 miles on the second part of the trip at a speed 5 mph slower. The total time

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: During the first part of a trip a canoeist travels 48 miles at a certain speed. The canoeist travels 19 miles on the second part of the trip at a speed 5 mph slower. The total time      Log On


   

Question 401141: During the first part of a trip a canoeist travels 48 miles at a certain speed. The canoeist travels 19 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hours. What was the speed on each part of the trip?
Found 2 solutions by jsmallt9, robertb:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Using the distance equals rate times time, d = r*t, we can write an equation for each part of the trip:
d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
with the subscripts describing the part of the trip. Since we know the distances for each part we can substitute the numbers for those distaces:
48+=+r%5B1%5D%2At%5B1%5D
19+=+r%5B2%5D%2At%5B2%5D

At this point we have two equations and four variables. (r%5B1%5D and r%5B2%5D count as separate variables. So do t%5B1%5D and t%5B2%5D.) So we need two more equations. We are told that the rate during the second part was 5 mph slower than the rate during the first part. So:
r%5B2%5D+=+r%5B1%5D+-+5
We are also told that the total time was 3 hours. So:
t%5B1%5D+%2B+t%5B2%5D+=+3
If we subtract t%5B1%5D from each side we can solve for t%5B2%5D:
t%5B2%5D+=+3+-+t%5B1%5D
Now we can use these last two equations and substitute for r%5B2%5D and t%5B2%5D in the 19+=+r%5B2%5D%2At%5B2%5D equation:
19+=+%28r%5B1%5D-5%29%2A%283-t%5B1%5D%29
which simplifies as follows:
19+=+r%5B1%5D%2A3-r%5B1%5D%2At%5B1%5D-5%2A3-5%2At%5B1%5D
19+=+3r%5B1%5D-r%5B1%5D%2At%5B1%5D-15-5%2At%5B1%5D
From 48+=+r%5B1%5D%2At%5B1%5D we know that r%5B1%5D%2At%5B1%5D is 48. Replacing the r%5B1%5D%2At%5B1%5D with 48 in the previous equation we get:
19+=+3r%5B1%5D-48-15-5%2At%5B1%5D
or
19+=+3r%5B1%5D-63-5%2At%5B1%5D
Adding 63 to each side we get:
82+=+3r%5B1%5D-5%2At%5B1%5D
Next we can solve for t%5B1%5D in the equation 48+=+r%5B1%5D%2At%5B1%5D by dividing both sides by r%5B1%5D:
48%2Fr%5B1%5D+=+t%5B1%5D
Now we can substitute for t%5B1%5D in 82+=+3r%5B1%5D-5%2At%5B1%5D:
82+=+3r%5B1%5D-5%2A%2848%2Fr%5B1%5D%29
We finally have an equation with a single variable! We can now solve for r%5B1%5D. Multiplying both sides by r%5B1%5D (to eliminate the fraction) we get:
82r%5B1%5D+=+3r%5B1%5D%5E2-240
This is a quadratic equation so we want one side to be zero. Subtracting 82r%5B1%5D from each side we get:
0+=+3r%5B1%5D%5E2-82r%5B1%5D-240
There are just too many possible factors of 240 for me to want to try to factor this. So I will use the Quadratic Formula instead:
r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%28%28-82%29%5E2-4%283%29%28-240%29%29%29%2F2%283%29
which simplifies as follows:
r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%286724-4%283%29%28-240%29%29%29%2F2%283%29
r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%286724%2B2880%29%29%2F2%283%29
r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%289604%29%29%2F2%283%29
r%5B1%5D+=+%2882+%2B-+sqrt%289604%29%29%2F6
r%5B1%5D+=+%2882+%2B-+98%29%2F6
In long form this is:
r%5B1%5D+=+%2882+%2B+98%29%2F6 or r%5B1%5D+=+%2882+-+98%29%2F6
which simplify as:
r%5B1%5D+=+%28180%29%2F6 or r%5B1%5D+=+%28-16%29%2F6
r%5B1%5D+=+30 or r%5B1%5D+=+%28-8%29%2F3
(BTW, this means the equation would have factored into:
0+=+%28r%5B1%5D-30%29%283r%5B1%5D%2B8%29

The second solution for the rate is negative which makes no sense in the context of this word problem. So we will reject it.

So r%5B1%5D, the rate during the first part of the trip, is 30 mph. The rate during the second part, r%5B2%5D, is 5 mpg slower so it must be 30-5 or 25 mph. (Since the problem only asks for the rates we are finished. We could use these values to find t%5B1%5D and t%5B2%5D also.)

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let r = speed at the 1st part of trip ==> time for 1st part = 48/r
==> r - 5 = speed at 2nd part of trip ==> time for 2nd part = 19/(r-5).
both from the formula t = d/r = distance/rate.
Then 48%2Fr+%2B+19%2F%28r-5%29+=+3
<==> 48(r-5) + 19r = 3r(r-5)
<==> 48r+-+240+%2B+19r+=+3r%5E2+-+15r
<==> 3r%5E2+-+82r+%2B+240+=+0 <==> (r - 24)(3r - 10) = 0
==> r = 24, 10/3
Eliminate r = 10/3, because it would make r - 5 negative.
Hence the speed in the 1st part is 24 mph, and the speed in the 2nd part is 19 mph.