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During the first part of a rip a canoeist travels 43 miles at a certain speed. The canoeist ravels 19 miles on the seond part o...Asked by Lois
During the first part of a trip, a canoeist travels 93 miles at a certain speed. The canoeist travels 19 miles on the second part of the trip at a speed of 5 mph slower. The total time for the trip is 3 hrs. What was the speed on each part?
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Answered by
Reiny
let the "certain" speed be x mph
so time at the certain speed = 93/x
slower speed = x-5
time at slower speed = 19/(x-5)
so 93/x + 19/(x-5) = 3
times x(x-5)
93(x-5) + 19x = 3x(x-5)
93x - 465 + 19x = 3x^2 - 15x
3x^2 - 127x + 465 = 0
x = (127 ± √(10549)/6 = 38.28 or 4.045 mph
the "certain" speed of 4.045 mph would give us a negative speed for the second leg, so we have to reject that.
so that leaves us with a speed of 38.28 mph for the first part, and 33.28 mph for the second part of the trip.
(notice that 93/38.28 + 19/33.28 = 3)
Even though the answer fits the given data, the question is totally ridiculous.
I am an avid canoist and to be able to canoe at 38 mph is totally absurd.
so time at the certain speed = 93/x
slower speed = x-5
time at slower speed = 19/(x-5)
so 93/x + 19/(x-5) = 3
times x(x-5)
93(x-5) + 19x = 3x(x-5)
93x - 465 + 19x = 3x^2 - 15x
3x^2 - 127x + 465 = 0
x = (127 ± √(10549)/6 = 38.28 or 4.045 mph
the "certain" speed of 4.045 mph would give us a negative speed for the second leg, so we have to reject that.
so that leaves us with a speed of 38.28 mph for the first part, and 33.28 mph for the second part of the trip.
(notice that 93/38.28 + 19/33.28 = 3)
Even though the answer fits the given data, the question is totally ridiculous.
I am an avid canoist and to be able to canoe at 38 mph is totally absurd.
Answered by
Lois
Thank you so much for your help, and I would be interested in watching someone try to paddle a canoe or any boat that fast. Bet their arms would be very tired before they went 1/4 mile.
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