SOLUTION: During the first part of a​ trip, a canoeist travels 58 miles at a certain speed. The canoeist travels 6 miles on the second part of the trip at a speed 5 mph slower. The tot

Algebra ->  Length-and-distance -> SOLUTION: During the first part of a​ trip, a canoeist travels 58 miles at a certain speed. The canoeist travels 6 miles on the second part of the trip at a speed 5 mph slower. The tot      Log On


   

Question 1057405: During the first part of a​ trip, a canoeist travels 58 miles at a certain speed. The canoeist travels 6 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?
Answer by jorel555(1290) About Me  (Show Source):
You can put this solution on YOUR website!
Let s be the speed of the canoe on the first part of the trip. Then the second part would be s-5. So:
58/s + 6/s-5=3
58(s-5)+6s=3sē-15s
3sē-79s+290=0
Using the quadratic formula, i.e. x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ , we get two roots of 4.40913022972 and 21.9242031036. So the only choice that makes sense is the answer of 21.9242 mph for the first part, and 16.9242 mph for the second part. ☺☺☺☺
-
-
"Airplane A travels 2800 km at a certain speed. Plane B travels 2000 km at a speed 50 km/h faster than plane A in 3 hrs less time. Find the speed of each plane."?
-
Let s be the speed of the slower plane. Then the faster plane flies at s+50. So:
2800/s=2000/(s+50) + 3
2800(s+50)=2000s+3sē+150s
3sē-650s-140000=0
Factoring this, we get
(3s+400)(s-350)
s=-400/3 or 350
The slower plane flies at 350 kph, the faster plane at 400 kph. ☺☺☺☺