A Train Breaks DownDate: 06/30/2001 at 22:28:05 From: Steve Subject: A train question Hi there, I can't figure out how to do this. I know there's probably some sort of algebraic way around it, but I can't figure it out. Here it is: A train breaks down an hour after starting its journey. The engineer takes half an hour to make repairs, but the train can only continue at half its original speed and arrives at its destination two hours late. If the delay had occurred 100km farther on in the journey, the train would only have been one hour late. What was the distance of the journey? Date: 07/01/2001 at 14:32:36 From: Doctor Jaffee Subject: Re: A train question Hi Steve, Here is how I solved this problem. Draw a segment representing the path of the train from the beginning to the end of the trip. Let d = the distance of the trip let r = the average rate of speed let t = the time it takes to finish the trip Now, draw another segment representing the trip when the train breaks down after 1 hour. Put an X on the track where you think the breakdown occurs. Since d = rt and the train has been travelling 1 hour, the distance that the train has travelled so far is d = r(1) = r. After the repairs, the train travels at a rate of r/2 and the time it travels is 2 hours more than the trip would have been without the repairs, less the time it took before the train restarted, which is 1 1/2 hours. So the length of time for the second part of the trip is t + 2 - 1 1/2 = t + 1/2. The distance the train travels in the second half of the trip is (r/2)(t + 1/2). So, adding the distances of the two parts, we get r + (r/2)(t + 1/2) = d You can simplify this equation, replace rt with d and end up with an equation in terms of d and r. Now make a third line and plot the trip with the repairs being made 100 km down the track. I suggest you consider three parts of the trip: First, the train travels r km in 1 hour, then the train travels 100 km at which time it breaks down. Finally it travels the remainder of the trip at r/2 km/hr. Give it a try and if you want to check your solution with me, write back. If you are having difficulties, let me know and show me what you have done so far, and I'll try to help you some more. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
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