20030830, 14:18  #1 
Aug 2003
Snicker, AL
7·137 Posts 
colliding trains
There is a train track completely around a planet at its equator. This track is exactly 25,000 km long. Two trains pull out of the station travelling in opposite directions on the track. The first train leaves at 30 kph and the second train leaves at 25 kph. Each train accelerates by 1 khp for each hour it travels and the acceleration is steady meaning that after 1/2 hour the train is travelling 1/2 kph faster than it was.
These two trains are explosive. For each kph of their final velocity they will blast a crater one more meter deep at the point of impact. For example, if the trains meet with one travelling at 82 kph and the other at 67 kph, the combined velocity will blow a crater 149 meters deep. The task is to find out where on the globe they will meet, how fast will they be travelling at the time, and how long after departure until their catastrophic meeting. Where they meet should be expressed in Kilometers travelled, when in hours/minutes/seconds, and size of blast crater in meters. Fusion the obfuscator 
20030830, 15:49  #2 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
The trains meet at T=0:00:00 at the station and form a blast crater of 55 meters.
If you ignore that meeting and the fact that they would blow up at the station, their next meeting is within the next week on the opposite side of the planet. (For a more accurate answer, please consult a middle school math student.) 
20030830, 19:19  #3  
Aug 2002
Richland, WA
10000100_{2} Posts 
Quote:


20030830, 19:35  #4 
Aug 2003
Upstate NY, USA
101000110_{2} Posts 
WARNING: Posting possible answer!
I find through integration that time will be ~132:59:15.13, the crash will be located ~12167.53115km from the station (the faster train covers ~12832.46885km) and the created blast has a radius of ~320.9750769m. 
20030830, 19:52  #5  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
You can derive x(t) = x0 + v0*t + 1/2 * a * t^2 without calculus. 

20030830, 19:54  #6 
Aug 2002
Richland, WA
84_{16} Posts 
That answer is the same as mine, but in the future you should PM (private message) your answers to the person who posted the puzzle to avoid giving away the answer too soon.

20030830, 20:50  #7 
Aug 2003
Snicker, AL
959_{10} Posts 
Wackerbarth,
I have to agree that this one is basic. My daughter was solving similar equations in high school algebra 2 years ago. There is a place on this forum for just such questions though. What bothers me is that 9 out of 10 high school graduates could not solve this equation. Fusion 
20030830, 22:35  #8 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
I find no fault with the problem except that I would have preferred it if you had chosen a slightly different circumference so that the answer came out "even".
As for my answer, I was intentionally avoiding giving too many details in hopes that others would also go through the exercise before the answer was disclosed. 
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