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

Algebra ->  Square-cubic-other-roots -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Let s = canoeist's speed for 1st part of trip in mi/hr
Let t = time in hrs to go 55 mi on 1st part of trip
+5+-+t%0D%0Agiven%3A%0D%0A1st+part%3A%0D%0A%281%29+%7B%7B%7B+55+=+s%2At+
2nd part:
(2) +4+=+%28+s+-+5+%29%2A%28+5+-+t+%29+
-------------------------------
(1) +t+=+55%2Fs+
(2) +4+=+5s+-+25+-+s%2At+%2B+5t+
Substitute (1) into (2)
(2) +4+=+5s+-+25+-+55+%2B+5%2A%2855%2Fs%29+
Multiply both sides by s
(2) +4s+=+5s%5E2+-+80s+%2B+275+
(2) +5s%5E2+-+84s+%2B+275+=+0+
Use quadratic equation:
+s+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+5+
+b+=+-84+
+c+=+275+
+s+=+%28-%28-84%29+%2B-+sqrt%28+%28-84%29%5E2+-+4%2A5%2A275+%29%29%2F%282%2A5%29+
+s+=+%28+84+%2B-+sqrt%28+7056+-+5500+%29%29%2F+10+
+s+=+%28+84+%2B-+sqrt%28+1556+%29%29%2F+10+
+s+=+%28+84+%2B-+39.4462%29%2F+10+
+s+=+%28+84+%2B+39.4462+%29+%2F+10+
+s+=+123.4462+%2F+10+
+s+=+12.3446+
+s+-+5+=+7.3446+
check answers:
(1) +55+=+s%2At+
(1) +55+=+12.3446t+
(1) +t+=+4.4554+ hrs
and
(2) +4+=+%28+s+-+5+%29%2A%28+5+-+t+%29+
(2) +4+=+%28+s+-+5+%29%2A%28+5+-+4.4554+%29+
(2) +4+=+%28+s+-+5+%29%2A.5446+
(2) +s+-+5+=+7.3448+
(2) +s+=+12.3448+
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) About Me  (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
sqrt%28%091556%09%29= 39.45 5
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
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