Question 939601: A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 of its former rate and arrives three hour after the time; but had the accident happened 50 miles farther on yhe line, it would have arrived one and one-half hour sooner. Find the length of the journey.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 of its former rate and arrives three hour after the time;
but had the accident happened 50 miles farther on the line, it would have arrived one and one-half hour sooner.
Find the length of the journey.
:
Let s = normal speed of the train
then
.6s = speed after accident
:
d = length of journey
then
 = normal travel time for journey
:
+ 2 = travel time with accident (don't count the 1 hr delay)
1s = distance traveled before the accident
normal time + 2 hrs = 1 hr + time at .6s
d/s + 2 = 1 + (d-1s)/.6s
multiply equation by .6s, cancel the denominators
.6d + 1.2s = .6s + d - s
.6d + 1.2s = -.4s + d
1.2s + .4s = d - .6d
1.6s = .4d
divide both sides by .4
d = 4s
:
d/s + 1.5 hr = travel time if accident happened 50 mi further
d/s + 1.5 = 1 + 50/s + (d - 1s - 50)/.6s
subtract 1 from both sides
d/s + .5 = 50/s + (d - 1s - 50)/.6s
multiply equation by .6s
.6d + .6s(.5) = .6(50) + d - 1s - 50
.6d + .3s = 30 + d - 1s - 50
.3s + 1s = d -.6d - 20
1.3s = .4d - 20
replace d with 4s
1.3s = .4(4s) - 20
1.3s = 1.6s - 20
20 = 1.6s - 1.3s
20 = .3s
s = 20/.3
s = 66.67 is the normal speed
find d
d = 4(66.67)
d = 266.667 mi is the distance of the journey
:
Check this using the 1st scenario\
time at normal speed, 1 hr
time at slower speed: (266.667-66.667)/.6(66.667) = 200/4 = 5 hrs
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total time 6 hrs
Find the time with no accident, normal arrival
266.667/66.667 = 4 hrs
Accident time was 2 hrs longer like it said
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