During the first part of a trip a canoeist travels 54 miles at a certain speed. The canoeist travels 20 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 4 hours. What was the speed on each part of the trip?

Question

owlfred · Answer

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dumbcow · Answer

d=r*t  (d is distance, r is speed, t is time)
First part:
54 = r*t1
second part:
20 = (r-5)*t2
t1 + t2 = 4
system of 3 equations with 3 variables
use substitution
t1 = 4-t2
r = 54/t1 => t1 = 54/r
54/r = 4-t2 => t2 = 4-54/r
$$20 = (r-5)(4-\frac{54}{r})$$
$$20 = 4r +\frac{270}{r} -54-20$$
$$4r -94 + \frac{270}{r} = 0$$
multiply everything by r
$$4r^{2}-94r +270 = 0$$
quadratic formula
$$r = \frac{94 \pm 67.2}{8}$$
$$r = 3.35 , 20.15$$
since you cant go neg speed, r>5 thus r=20.15
2 speeds:
20.15 and 15.15

dumbcow · Answer

were you able to follow and understand how i got the answers?

anonymous · Answer

yes I was thank you

dumbcow · Answer

:)