SOLUTION: I need help, I have tried this question several times and still am not getting it right. During the first part of a trip, a canoeist travels 55 miles at a certain speed. The ca

Question 574504: I need help, I have tried this question several times and still am not getting it right.
During the first part of a trip, a canoeist travels 55 miles at a certain speed. The canoeist travels 4 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 5 hrs. What was the speed on each part of the trip?
Found 2 solutions by josmiceli, mananth:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = canoeist's speed for 1st part of trip in mi/hr
Let = time in hrs to go 55 mi on 1st part of trip

2nd part:
(2)
-------------------------------
(1)
(2)
Substitute (1) into (2)
(2)
Multiply both sides by
(2)
(2)
Use quadratic equation:

check answers:
(1)
(1)
(1) hrs
and
(2)
(2)
(2)
(2)
(2)
This looks close enough
The speed for the 1st part was 12.3448 mi/hr
The speed for the 2nsd part was 7.3448 mi/hr

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
First part 55 miles
Second part 4 miles

First part x mph
Second part x -5 mph
Total time 5 hours
First part time 55 / x
Second part time 4 / ( x -5 )

Time first part + time second part = 5 hours

55 / x + 4 /(x -5 ) = 5
LCD = ( x + 0 )* (x -5 )
multiply the equation by the LCD
we get
55 * (x+ -5 )+ 4 x = 5
55 x+ -275 + 4 x = 5 X^2 -25 x
84 x+ -275 = 5 X^2
5 X^2 -84 x+ 275 = 0
5 X^2+ -84 x+ 275 = = 0
/ 5
5 X^2 -84 x+ 275 = 0

Find the roots of the equation by quadratic formula

a= 5 b= -84 c= 275

b^2-4ac= 7056 - 5500
b^2-4ac= 1556
= 39.45 5

x1=( 84 + 39.45 )/ 10
x1= 12.34
x2=( 84 -39.45 ) / 10
x2= 4.46
Ignore x2
x = 12.34 mph

CHECK
Time first part + Time second part
4.46 + 0.51 = 4.96

m.ananth@hotmail.ca

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