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, 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 + ({{{(d-s)/(.6s)}}}) - {{{d/s}}} = 3
2 + ({{{(d-s)/(.6s)}}}) - {{{d/s}}} = 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/.4}}}s
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/s}}} + ({{{(d-s-50)/(.6s)}}}) - {{{d/s}}} = 3 - 1.5
2 + {{{50/s}}} + ({{{(d-s-50)/(.6s)}}}) - {{{d/s}}} = 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