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train accident

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by maihuna » Thu Jun 04, 2009 4:48 am
A train, an hour after starting, mets with an accident which detains it for an hour, after which it proceeds at 3/5th of its former speed and arrives 3 hours after time: But had the accident happened 100 miles farther on the line, it would have arrived 1.5 hrs sooner. What is the approximate length of the journey?

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by sportcntr3 » Thu Jun 04, 2009 8:31 am
Kind of a shot in the dark here, but here it goes.

Find the initial speed of the train by figuring out what speed the train would need to travel in order to turn that extra 100 miles it gets in the second scenario into 1.5 hours of shorter time,

i.e. how fast must the train travel to go 100 miles in 1.5 hours:

100/1.5 = 66.6667

Initial speed of train = 66 2/3 mph
After Accident speed = 66 2/3 * 3/5 = 40 mph

D=R*T

D= 66 2/3 * 1hour + 40 * 3hours (ignoring the one hour layover) = 186 2/3 miles
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by SanjeevK » Thu Jun 04, 2009 5:57 pm
Here we have the time difference of 1.5 hours because of difference speed with which the train travels to cover 100 miles.
Let the initial speed of the train is v.
So we get 100/(3v/5) - 100/v = 1.5
this gives v = 400/9. Hence the initial speed of the train is 400/9
Now let the total time taken for the journey (assuming no accident) is t
Equating for distance
400t/9 = 400/9 + 80(t-2+3)/3

This yields t = 4 hours.
Hence the total journey is 400*4/9 = 178 miles approximately.
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