SOLUTION: During the first part of a trip, a canoeist travels 69 miles at a certain speed. The canoeist travels 2 miles on the second part of the trip at a speed 5 mph slower. The total time

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Question 162664: During the first part of a trip, a canoeist travels 69 miles at a certain speed. The canoeist travels 2 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hrs. What was the speed on each part of the trip?
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(6784) About Me  (Show Source):
You can put this solution on YOUR website!
Let time for 1st part = t hrs
Then the time for the 2nd part will be 3+-+thrs
The speed for the 1st part is s
The speed for the 2nd part is s+-+5 mi/hr
The distance for the 1st part was 69 mi
The distance for the 2nd part was 2 mi
(1) 69+=+s%2At
(2) 2+=+%28s+-+5%29%2A%283+-+t%29
2+=+3s+-+15+-+st+%2B+5t
2+=+3s+-+15+-+69+%2B+5t
3s+=+2+%2B+15+%2B+69+-+5t
3s+=+86+-+5t
Substituting (1) into this,
3s+=+-5%2A%2869%2Fs%29+%2B+86
Multiply both sides by s
3s%5E2+=+-345+%2B+86s
3s%5E2+-+86s+%2B+345+=+0
Use quadratic equation
s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+3
b+=+-86
c+=+345
s+=+%2886+%2B-+sqrt%28+86%5E2-4%2A3%2A345+%29%29%2F%282%2A3%29+
s+=+%2886+%2B-+sqrt%287396+-+4140%29%29%2F6
s+=+%2886+%2B-+sqrt%283256%29%29%2F6
s+=+%2886+%2B+57.06%29%2F6
s+=+%2886+-+57.06%29%2F6
I'll pick the answer that makes the most sense
s+=+23.84
and
s+=+4.82(I can't go 5 mph slower than this, so I'll pick
the 1st answer)
s+-+5+=+18.84
The canoeist went 23.84 mi/hr on the 1st part of the trip and
18.84 mi/hr on the 2nd part
check answer:
(1) 69+=+s%2At
(2) 2+=+%28s+-+5%29%2A%283+-+t%29
69+=+23.84t
t+=+2.89hrs
(2) 2+=+18.84%2A%283+-+2.89%29
2+=+18.84%2A.11
2+=+2.07 (error is due to rounding off)
Hope I got it right!

Answer by ankor@dixie-net.com(12701) About Me  (Show Source):
You can put this solution on YOUR website!
During the first part of a trip, a canoeist travels 69 miles at a certain speed. The canoeist travels 2 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hrs. What was the speed on each part of the trip?
;
Canoing that far in 3 hrs is a fantasy, but anyway:
:
Let s = speed on the 69 miles
then
(s-5) = speed on the last to miles of the trip
:
Write a time equation: Time = dist/speed
:
69 mi time + 2 mi time = 3 hrs
69%2Fs + 2%2F%28%28s-5%29%29 = 3
:
Multiply equation by s(s-5), results:
69(s-5) + 2s = 3s(s-5)
69s - 345 + 2s = 3s^2 - 15s
71s - 345 = 3s^2 - 15s
0 = 3s^2 - 15s = 71s + 345
A quadratic equation:
3s^2 - 86s + 345 = 0
Use the quadratic formula to solve this
The solution that makes sense is s ~ 23.84 mph for the 1st 69mi
and
23.84 - 5 = 18.84 mph for the last two mi
:
:
Check solution by finding the times at each speed
69/23.84 = 2.89 hrs
2/18.84 = .106 hr
----------------
total time 2.996 ~ 3 hrs
:
I suspect there may be a typo in this as written here, but the method should
help you.