SOLUTION: a train, an hour after starting, meets with an accident which detains it an hour .after which it proceed at 3/5 of its former rate arrives 3hrs after time but had the accident happ

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Question 728865: a train, an hour after starting, meets with an accident which detains it an hour .after which it proceed at 3/5 of its former rate arrives 3hrs after time but had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner.find the length of the journey
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A train, an hour after starting, meets with an accident which detains it an hour.
After which it proceed at 3/5 of its former rate, arriving 3 hrs late.
But had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner.
Find the length of the journey
:
let s = normal speed of the train (s also = the dist traveled the 1st hr)
then
.6s = speed after the accident
:
let d = distance of the journey
and
d/s = normal time of the journey
:
Write a time equation for the 1st scenario
:
normal speed time + stopped time + 3/5 speed time - normal time = 3 hrs
1 + 1 + (%28d-s%29%2F%28.6s%29) - d%2Fs = 3
2 + (%28d-s%29%2F%28.6s%29) - d%2Fs = 3
multiply by .6s, resulting in:
2(.6s) + d - s - .6d = .6s(3)
1.2s + d - s - .6d = 1.8s
combine like terms
d - .6d = 1.8s - 1.2s + s
.4d = 1.6s
divide both sides by .4
d = 1.6%2F.4s
d = 4s
:
An equation for the 2nd scenario
" but had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner."
2 + 50%2Fs + (%28d-s-50%29%2F%28.6s%29) - d%2Fs = 3 - 1.5
2 + 50%2Fs + (%28d-s-50%29%2F%28.6s%29) - d%2Fs = 1.5
Multiply by .6s
.6s(2) + .6(50) + d - s - 50 - .6d = .6s(1.5)
1.2s + 30 + d - s - 50 - .6d = .9s
Combine like terms
1.2s - s - .9s + d -.6d -50 + 30 = 0
-.7s + .4d - 20 = 0
-.7s + .4d = 20
Replace d with 4s
-.7s + .4(4s) = 20
-.7s + 1.6s = 20
.9s = 20
s = 20/.9
s = 22.2 mph
then
4(22.2) = 88.8 mi is the journey