Bridge and Train Puzzle
July 4th, 2012 at 1:15:34 PM
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You are 40% of the way across a train bridge when you hear a train whistle behind you. You must choose to turn around or keep going forward to get off the bridge before the train runs you over. It turns out that regardless of which way you choose you will get off the bridge just in the knick of time. You can run 10 miles per hour. How fast is the train going?
As usual, please put solutions in.
As usual, please put solutions in
John Galt drives a Prius
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 4th, 2012 at 1:28:08 PM
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Quote: WizardYou are 40% of the way across a train bridge when you hear a train whistle behind you. You must choose to turn around or keep going forward to get off the bridge before the train runs you over.
If the whistle, along with the attached train, is behind you, wouldn't it run you over sooner if you turned around?
Random guess:
40 mph
Donald Trump is a fucking criminal
July 4th, 2012 at 2:18:03 PM
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Quote: NareedIf the whistle, along with the attached train, is behind you, wouldn't it run you over sooner if you turned around?
Yes, if it were going to run over you both ways, that would happen sooner if you turned back. But depending on speeds and where you are on the bridge, there may be reversing answers on which direction is better. In this case, they are equally marginal.
X = length of bridge
t1=time until train reaches bridge
t2=time until train has crossed bridge
S=train speed
t1=(0.4X/10)
t2=(0.6X/10)
t2-t1=X/S
0.6X/10 - 0.4X/10 = 0.2X/10 = X/S
S =10/0.2 = 50 mph
t1=time until train reaches bridge
t2=time until train has crossed bridge
S=train speed
t1=(0.4X/10)
t2=(0.6X/10)
t2-t1=X/S
0.6X/10 - 0.4X/10 = 0.2X/10 = X/S
S =10/0.2 = 50 mph
July 4th, 2012 at 2:23:02 PM
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Quote: DocYes, if it were going to run over you both ways, that would happen sooner if you turned back. But depending on speeds and where you are on the bridge, there may be reversing answers on which direction is better. In this case, they are equally marginal.
Ok.
BTW, your spoiler label shows the result.
Donald Trump is a fucking criminal
July 4th, 2012 at 2:31:23 PM
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deleted
DUHHIIIIIIIII HEARD THAT!
July 4th, 2012 at 2:31:41 PM
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Quote: NareedOk.
BTW, your spoiler label shows the result.
I know. I thought we were supposed to hide the solution, i.e. the how to figure it out, not the answer. If I misunderstood, I apologize.
July 4th, 2012 at 5:08:32 PM
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Quote: DocI know. I thought we were supposed to hide the solution, i.e. the how to figure it out, not the answer. If I misunderstood, I apologize.
You're fine. I asked that only solutions be hidden. I figured nobody had to believe your answer, but I think in your case, Doc, everyone does.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 4th, 2012 at 5:15:32 PM
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Let L be the distance across the bridge you are when you hear the whistle. Since it is 40% of the way across the bridge, either you must turn around and run distance L, or keep going and run distance 1.5 L.
Let T be the distance the train is from the start of the bridge, and X be its velocity.
It takes you L/10 hours (if T is in miles) for you to reach the start of the bridge; it takes the train L/X hours (if L is in hours and X is in MPH) to get there. Therefore, L/10 = T/X.
It takes you 1.5 L/10 hours for you to reach the end of the bridge; it takes the train (T + 2.5L)/X hours. Therefore, 1.5 L/10 = (T + 2.5L)/X.
L/10 = T/X, so 1.5 L/10 = 1.5 T/X
1.5 T/X = (T + 2.5L)/X
1.5 T = T + 2.5 L
T = 5L
Substituting for T in L/10 = T/X, we get L/10 = 5L/X, so X = 50.
The train's velocity is 50 MPH.
Let T be the distance the train is from the start of the bridge, and X be its velocity.
It takes you L/10 hours (if T is in miles) for you to reach the start of the bridge; it takes the train L/X hours (if L is in hours and X is in MPH) to get there. Therefore, L/10 = T/X.
It takes you 1.5 L/10 hours for you to reach the end of the bridge; it takes the train (T + 2.5L)/X hours. Therefore, 1.5 L/10 = (T + 2.5L)/X.
L/10 = T/X, so 1.5 L/10 = 1.5 T/X
1.5 T/X = (T + 2.5L)/X
1.5 T = T + 2.5 L
T = 5L
Substituting for T in L/10 = T/X, we get L/10 = 5L/X, so X = 50.
The train's velocity is 50 MPH.
July 4th, 2012 at 5:28:29 PM
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L: Length of the bridge
You can get .4*L on the bridge before the train reaches said bridge. That means if you continue walking in the same direction as the train, you'll reach .8*L. In other words, in the time it takes you to run .2*L the train covers the entire length L... so the train is five times faster than you or 50 mph.
Did someone just watch Stand By Me?
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"
July 4th, 2012 at 5:52:12 PM
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Let bridge span be L, with L1 distance to the train.
Let train velocity be v.
Don't click the spoiler until reading the rest of the post.
I hoped the additional factor would play some role, or you could determine L, but it's as straightforward as it looks, I can't see any trick if there should be one.
Let train velocity be v.
Don't click the spoiler until reading the rest of the post.
At the moment you hear the whistle (t1), the train has traveled L1*v/c, c=340
At the moment you are being run over as you are exiting the bridge (t2), the train has traveled L1*v/340+v*.6L/4.4704
At the moment you run into the train as you backpedal, the train has traveled L1*v/340+v*.4L/4.4704
These are separated by the bridge: L1*v/340+v*.4L/4.4704+L=L1*v/340+v*.6L/4.4704
.4Lv/4.4704+L-.6Lv/4.4704=0
-.2Lv/4.4704=-L
Lv=L*4.4704/.2
v=22.352 (same answer)
At the moment you are being run over as you are exiting the bridge (t2), the train has traveled L1*v/340+v*.6L/4.4704
At the moment you run into the train as you backpedal, the train has traveled L1*v/340+v*.4L/4.4704
These are separated by the bridge: L1*v/340+v*.4L/4.4704+L=L1*v/340+v*.6L/4.4704
.4Lv/4.4704+L-.6Lv/4.4704=0
-.2Lv/4.4704=-L
Lv=L*4.4704/.2
v=22.352 (same answer)
I hoped the additional factor would play some role, or you could determine L, but it's as straightforward as it looks, I can't see any trick if there should be one.
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July 4th, 2012 at 7:01:11 PM
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ahiromu has the most elegant solution, in my opinion. It's also how i reasoned it out.
:P
:P
Wisdom is the quality that keeps you out of situations where you would otherwise need it
July 4th, 2012 at 7:59:23 PM
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same answer as above, but no math
If you run backward 40% of bridge, you and the train will both be at start of bridge.
If you run forward 40% of bridge, you will be 20% away from bridge end. The train will still be at the start of bridge.
So you are 20% away, and the train is 100% away.
Then the train is 5 times as fast as you. ie: 10 x 5 = 50 mph
If you run backward 40% of bridge, you and the train will both be at start of bridge.
If you run forward 40% of bridge, you will be 20% away from bridge end. The train will still be at the start of bridge.
So you are 20% away, and the train is 100% away.
Then the train is 5 times as fast as you. ie: 10 x 5 = 50 mph
September 22nd, 2013 at 8:28:14 AM
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Is this a contest for the right answer or the simplest answer? Here is a pretty simple one:
The length of the bridge should be immaterial, so let's call it one mile.
The runner and the train can get to the foot of the bridge in 0.04 hours.
The runner and the train can get to the end of the bridge in 0.06 hours.
Therefore the train can cross the bridge in 0.02 hours so its speed is 50 mph.
The runner and the train can get to the foot of the bridge in 0.04 hours.
The runner and the train can get to the end of the bridge in 0.06 hours.
Therefore the train can cross the bridge in 0.02 hours so its speed is 50 mph.
September 22nd, 2013 at 9:03:17 AM
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You can run 10 miles per hour, but do you have to? Imagine a bridge with a span of less than one stride. You are standing 40% of the way across, hear the train traveling at 1 mph. As it nears the bridge, you play a little "chicken", and just before it reaches you, you step off. Doesn't matter which direction, you just made it.
A falling knife has no handle.
September 22nd, 2013 at 12:37:22 PM
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Quote: MoscaYou can run 10 miles per hour, but do you have to? Imagine a bridge with a span of less than one stride.
A one stride bridge, where do you live, Mosca, Lilliputia?
(bet you have to look that up)
"It's not called gambling if the math is on your side."
September 22nd, 2013 at 12:59:43 PM
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Quote: EvenBobQuote: MoscaYou can run 10 miles per hour, but do you have to? Imagine a bridge with a span of less than one stride.
A one stride bridge, where do you live, Mosca, Lilliputia?
(bet you have to look that up)
Oh come on, Bob, you really think I would have to look that one up? The bridge could be right here, if the runner was Brobdingnabian!
I understand that the original problem is analog and algebraic, I was proposing a situation where it would break down because a stride represents a unit that would be greater than the distance in the problem. Sorry I don't have the math words to say it.
A falling knife has no handle.
September 22nd, 2013 at 12:59:54 PM
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So there are two unknowns: x = how far away is the train from you, and v = how fast is the train.
If you keep running: distance from the train position to the end of the bridge = velocity of train * time you need to run to the end of the bridge
x + 0.6 = v * 0.6 / 10
If you turn around: distance from the train position to the beginning of the bridge = velocity of train * time you need to run to the beginning of the brige
x - 0.4 = v * 0.4 / 10
substract both lines (since you don't care about x):
1 = v * 0.02 resulting in v = 50
If you keep running: distance from the train position to the end of the bridge = velocity of train * time you need to run to the end of the bridge
x + 0.6 = v * 0.6 / 10
If you turn around: distance from the train position to the beginning of the bridge = velocity of train * time you need to run to the beginning of the brige
x - 0.4 = v * 0.4 / 10
substract both lines (since you don't care about x):
1 = v * 0.02 resulting in v = 50
September 22nd, 2013 at 1:03:57 PM
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Quote: MoscaThe bridge could be right here, if the runner was Brobdingnabian!
That would be Mission, he's from Brobdingnabian, or so I hear.
"It's not called gambling if the math is on your side."
