SOLUTION: during the first part of a trip , a canoeist travels 37 miles at a certain speed. the canoeist travels 17 miles on the second part of the trip at a speed 5 mph slower. the total ti

Question 515160: during the first part of a trip , a canoeist travels 37 miles at a certain speed. the canoeist travels 17 miles on the second part of the trip at a speed 5 mph slower. the total time for the trip is 2 hrs. What was the speed on each part of the trip?
Found 2 solutions by drcole, lwsshak3:
Answer by drcole(72) (Show Source): You can put this solution on YOUR website!
Let s represent the speed of the canoeist during the first part of the trip (in miles per hour). Since the speed is constant over the first part of the trip, we can use the relationship
Distance = Speed * Time
to solve for the time it took for the canoeist to travel the first leg in terms of s. In this case, the distance was 37 miles, so we have the equation
37 = s * Time
= Time (dividing both sides by s, presuming that s is not 0, which is reasonable since the canoeist traveled 37 miles at that speed)
So now we have an expression for the amount of time it took to travel the first leg: . Now we move on to the second leg. According to the question, the canoeist traveled at a speed 5 mph slower than during the first leg of the trip. We can therefore represent the speed during the second leg as . Now we'll solve for the time it took to travel the second leg. The same relationship between distance, speed, and time applies, but now the distance traveled is 17 miles and the speed is mph. Solving for time, we get:
17 = * Time
= Time (dividing both sides by , which we can also assume is non-zero since it is the speed of the canoeist)
The final bit of information we are given is that the total time of the trip is 2 hours. We now have the times it took to travel the first and second legs of the trip in term of s, and we can set up an algebraic equation:

Now we solve for s:
(multiply both sides by s)
(multiply both sides by )
(distribute the 37 on the left hand side and the on the right)
(combine like terms)
(subtract 54s from and add 185 to both sides)
(applying the quadratic formula)
(simplifying)
So we have two possible solutions for the speed. Numerically, these two solutions are (to the nearest hundredth):
mph and mph
Both are positive, so both might be speeds for the first leg of the trip. We know, however, that the speed during the second leg of the trip was , so:
if mph, then mph; and
if mph, then mph.
Speed must be a non-negative number, so -1.79 mph cannot be a speed for the second leg of the trip. Therefore we reject the second solution, giving us that the canoeist traveled at approximately 28.79 mph during the first part of trip, and approximately 23.79 mph during the second part of the trip. To check, note that the time it took to travel the first leg would therefore be (to the nearest hundredth):
hours
and the time it took to travel the second leg would be (again, to the nearest hundredth):
hours
making the total time traveled 1.29 + 0.71 = 2.00 hours, which is what we wanted.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
during the first part of a trip , a canoeist travels 37 miles at a certain speed. the canoeist travels 17 miles on the second part of the trip at a speed 5 mph slower. the total time for the trip is 2 hrs. What was the speed on each part of the trip
**
let x=speed during first part of trip
x-5 =speed during second part of trip
travel time=distance/speed
17/(x-5)+37/x=2
LCD: x(x-5)
17x+37(x-5)=2x(x-5)
17x+37x-185=2x^2-10x
2x^2-64x+185=0
solve by following quadratic formula
..

..
a=2, b=-64, c=185
x=[-(-64)�√(64^2-4*2*185)]/2*2
x=[64�√(4096-1480)]/4
x=[64�√2616]/4
x=(64�51.15)/4
x=28.79
or
x=3.21 (reject, (3.21-5)<0) 2nd part of trip
ans:
speed during first part of trip=28.79 mph
speed during second part of trip=23.79 mph
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