During the first part of a trip, a fisherman travels 45 miles at a certain speed… what was the total speed?
05.September, 2009
During the first part of a trip, a fisherman travels 45 miles at a certain speed. The fisherman travels 10 miles on the second part of the trip at a speed of 5mph slower. The total time of the trip was 3 hours. What was the total time for each part of the trip?
First part of trip?
Second part?
t = first part of trip
3-t = second part of trip
v = speed of the first part
v-5 = speed of the second part
45 = v*t……(1)
10 = (v-5)*(3-t) => 10+5(3-t) = v(3-t)……(2)
(2)/(1): (25-5t)/45 = (5-t)/9 = (3-t)/t
Cross mulitplication: 5t- t^2 = 27-9t
=> t^2-14t+27 = 0
Solving with quadratic formula,
t = 2.3096 hours
3-t = .6904 hours
05.September, 2009 um 4:25 pm
t = first part of trip
3-t = second part of trip
v = speed of the first part
v-5 = speed of the second part
45 = v*t……(1)
10 = (v-5)*(3-t) => 10+5(3-t) = v(3-t)……(2)
(2)/(1): (25-5t)/45 = (5-t)/9 = (3-t)/t
Cross mulitplication: 5t- t^2 = 27-9t
=> t^2-14t+27 = 0
Solving with quadratic formula,
t = 2.3096 hours
3-t = .6904 hours
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