During the first part of a trip, a canoeist travels 96 miles at a certain speed. the canoeist travels 15 miles on the second part of the trip at a speed 5 mph slower. the total time for the trip is 4 hrs. what was the speed on each part of the trip?
Accepted Solution
A:
This problem is a problem using d = r * t First part of the trip d = 96 miles r = ?? t = ??
Second part of the trip d = 15 miles r = r - 5 t = 4 - t
Solve for r first. 96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5 96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left 96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left. 111r - 480 = 4r(r - 5) Remove the brackets on the right. 111r - 480 = 4r^2 - 20r Bring the left side over to the right side. 0 = 4r^2 - 20r - 111r + 480 Collect like terms. 0 = 4r^2 - 131r + 480
The only way I could do this was to use the quadratic equation. [tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} [/tex] a = 4b = - 131c = 480[tex]\text{x = }\dfrac{ -(-131) \pm \sqrt{(-131)^{2} - 4*4*480 } }{2*4} [/tex] [tex]\text{x = }\dfrac{ 131 \pm \sqrt{(17161) - 7680 } }{8} [/tex] [tex]\text{x = }\dfrac{ 131 \pm \sqrt{9481 } }{8} [/tex] [tex]\text{x = }\dfrac{ 131 \pm 97.37 }{8} [/tex]
x = (131 + 97.38)/8 x = 28.55 x = (131 - 97.37)/8 x = 4.20
4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.
r = 28.55 for the first part of the trip and r = 23.55 for the second part of the trip.
Check. Let's see if the times add up to 4 d/r = t 96/28.55 = 3.36 15/22.75 = 0.64
The times do = 4, so the rates are correct.
Answers r1 = 28.55 for the 96 mile trip. r2 = 23.55 for the 15 mile trip.