Question 30584: A train, an hour after starting, meets an accident which detains it one hour, after which it proceed at 3/5 of its former rate and arrives 3 hours late. But, had accident happened 80km farther, it would have been 1.5 hours late only. Find the original rate of the train and the length of the journey.
Answer by Fermat(136)   (Show Source):
You can put this solution on YOUR website! Let v1 be the speed before the breakdown
Let v2 be the speed after the breakdown
v2 = (3/5)v1
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The two journeys are identical except for the 80km part
let s1 be the distance up to the 80 km part
The difference in time taken between travelling the 80 km at v1 and at v2 is equal to the difference in time of the two late journeys - which difference is 1.5 hrs.
In other words, it takes 1.5 hrs longer to travel 80 km when trvelling at v2 than it does when travelling at v1.
time to travel 80 km at v1 = 80/v1 = t1
time to travel 80 km at v2 = 80/v2 = t2
t2 - t1 = 1.5
80/v2 - 80/v1 = 3/2
80(1/v2 - 1/v1) = 3/2
(80/v1)(5/3 - 1) = 3/2
(80/v1)(2/3) = 3/2
80/v1 = 9/4
v1 = 320/9
v1 = 35.5556 km/hr
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Let s2 be the distance travelled after the breakdown (in the first joiuurney), such that,
s1 + s2 = L, where L is the overall journey distance travelled from the start to the final destination.
Let T be the journey time without breakdowns.
L = v1T
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In the first journey, excluding the 1hrs wait time, it takes 2hrs longer to travel the distance s2, at the speed v2, than at the speed v1 (normal/standard speed)
t = s2/v1
t' = s2/v2
t' - t = 2
s2(1/v2 - 1/v1) = 2
(s2/v1)(5/3 - 1) = 2
s2/v1 = 2*3/2 = 3
s2 = 3v1
s2 = 106.6667 km
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The distance s1 is travelled in 1 hr (given) at speed v1
s1 = 35.5556*1
s1 = 35.5556 km
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L = s1 + s2
L = 35.5556 + 106.6667
L = 142.2222 km
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