| Member |
 Discussion Topic  |
|
|
Deseret Rider
 Advanced Member
776 Posts
[Mentor]
Helper, Utah
USA
BMW
R1100RT
|
 Posted - 12/15/2009 : 11:15 PM
Like
|
Two trains, each two miles long, enter two one mile long tunnels that are two miles apart from one another on the same track. The trains enter the tunnels at exactly the same time. The first train is going 5 miles/hour, and the second train is going 10 miles/hour. What is the sum of the lengths of the two trains that will protrude from the tunnels at the exact moment that they collide, assuming that neither train changes its speed prior to collision? The trains are on the same track headed in opposite directions (i.e. directly toward one another).
|
|
|
Daddio
Advanced Member
775 Posts
[Mentor]
Calera, AL
USA
Suzuki
Bandit 1250
|
Posted - 12/16/2009 : 12:01 AM
|
Total length of trains = 4 miles
Total length of tunnels = 2 miles
Total length of trains protruding = length of trains - length of tunnels
Total length of trains = 4 - 2 = 2 miles
The trains will meet somewhere before either can enter the other tunnel. My estimate puts the fast train 1 2/3 miles into the gap between the tunnels. The slow train 1/3 mile into the gap.
Regardless you hve 4 miles of train covered by two miles of tunnel. 2 miles is the sum of protruding train.
edit I am wrong - you will have 1 1/3 mile covered by tunnel. 2 2/3 mile is the sum. Final answer, promise. |
Edited by - Daddio on 12/16/2009 12:20 AM |
 |
|
|
scottrnelson
Advanced Member
6943 Posts
[Mentor]
Meridian, ID
USA
Honda
XR650L, 790 Adv R
|
Posted - 12/16/2009 : 7:19 AM
|
The one train is going twice as fast as the other one. It doesn't matter what the actual speeds are. That means the faster train covers twice the distance in the same time, so the slow train covers 1/3 of the 4 miles and the fast train covers 2/3 of the four miles. They meet when the slow train has gone 4/3 = 1 1/3 miles and the fast train has gone 8/3 = 2 2/3 miles.
Since the slow train has both ends sticking out of the 1 mile tunnel, it has exactly 1 mile exposed.
The fast train has gone 5/3 = 1 2/3 from the exit of its tunnel, so that is how much is exposed.
Total exposed train is 5/3 + 3/3 = 8/3 = 2 2/3 miles of train exposed.
So it looks like Daddio's revised answer was right. |
|
|
|
Deseret Rider
Advanced Member
776 Posts
[Mentor]
Helper, Utah
USA
BMW
R1100RT
|
Posted - 12/17/2009 : 9:50 PM
|
Good work both of you. Your thought process being a little different both arrived at the same answer I have. My line of reasoning was very lengthy but after much effort to simplify this is the way it came down for me.
2 2/3 miles. The trains are exactly 4 miles apart. Their combined speed is 15 miles/hour, so it will take them 16 minutes to collide. The first train will have travelled 1 1/3 miles, so 1/3 mile of it will be out of the tunnel in front, & 2/3 mile of it will be out in the back on collision. The other mile of it will be in the tunnel. The second train will have travelled 2 2/3 miles, so only 1/3 mile of it will still be in the tunnel, so 1 2/3 miles of it will be out of the tunnel. By addition then 2 and 2/3rds miles of trains are exposed out of the tunnel. |
|
|
|
scottrnelson
Advanced Member
6943 Posts
[Mentor]
Meridian, ID
USA
Honda
XR650L, 790 Adv R
|
Posted - 12/18/2009 : 6:42 AM
|
quote:
Originally posted by Deseret Rider
Your thought process being a little different both arrived at the same answer I have.
After reading Surely You're Joking, Mr. Feynman about the physicist Richard P. Feynman, and finding out that he did math differently than others, I've stopped trying to conform to "the correct way" of computing things. I wish I had figured that out way back in elementary school. |
|
|
| |
Discussion Topic |
|
The trains and tunnels math problem