Need to understand the logic behind: 2 Trains

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Dear all,
following question may seem very easy for some, but I hope to get an answer, because I really want to understand why following is done:
Train A and B are travelling from opposite ends of 500mile long track at 1 pm.
Train A travells 35m/h, Train B 25/m. At what time will they meet?

I know the principle of: Distance = Rate * Time

But I dont understand (= it does not seem logic) why the answer says that I need to combine the rates together, as the trains come from opposite site.

Any clear explanation is highly valued!
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by sl750 » Tue Oct 04, 2011 11:25 pm
You are using the concept of relative velocity

When two objects are traveling in the same direction,then the difference in speed between their individual speed is their relative speed . |R1-R2|

When two objects are traveling in opposite direction, then the sum of their individual speed is the their relative speed. R1+R2

Hence, in your example, you add the individual speeds

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by svd.kumar » Wed Oct 05, 2011 7:03 am
Hi

Could you please give the answer, for me I got it as 9:20 PM.

Thanks,
SVD

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by Abhishek009 » Wed Oct 05, 2011 9:37 am
PhDmessi wrote:Dear all,
following question may seem very easy for some, but I hope to get an answer, because I really want to understand why following is done:
Train A and B are travelling from opposite ends of 500mile long track at 1 pm.
Train A travells 35m/h, Train B 25/m. At what time will they meet?

I know the principle of: Distance = Rate * Time

But I dont understand (= it does not seem logic) why the answer says that I need to combine the rates together, as the trains come from opposite site.

Any clear explanation is highly valued!
I hope this might help U :

https://mathforum.org/dr.math/faq/faq.two.trains.html
Abhishek

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by Brian@VeritasPrep » Thu Oct 06, 2011 10:24 am
Hey PhD,

Great question - this is such a classic rate problem setup, immortalized in American sitcoms starring frustrated high school students!

You don't **have to** add the rates, although it's pretty helpful. Another way to look at this is to note that you have two rates:

Rate(A) = Distance A / Time A

Rate(B) = Distance B / Time B

And they give us rate A and rate B:

35 = Distance A / Time A
25 = Distance B / Time B

Now, the problem here is that we have two equations and four variables. So we need to find relationships between those variables so that we can solve for them...if we have 2 equations, 2 variables, we can do it, but 2 equations & 4 variables we can't.

What works in this problem is that we know that the distances and times are not independent:

Distance
Whatever distance train A travels, train B doesn't have to travel. They combined have 500 miles to cover, but if A goes 200 then B only needs to go 300. So the distance of train B is just the portion of the 500 that train A doesn't cover. So D(B) = 500 - D(A).

This, also, is why you can combine the rates. They're working together to cover the same 500-mile distance, so while one is going 35 miles/hour from one end, the other is going 25 miles/hour. So every hour, we chop off 35 miles from one end and 25 miles from the other end; every hour, they're 60 miles closer to meeting.

Time
Two trains/people/items can only "meet" at the same time. If I'm at Starbucks at 14th and Broadway at 2pm on a Wednesday and you're there at 3pm on a Friday, we won't meet. But if we're both there at noon on Thursday, we will. So in these problems there's always a relationship between the times...two things can only meet at the same time. And since they both left at the same time, their times are equal (had one left an hour earlier, it would travel an hour longer, or T+1). So T(A) = T(B).

Putting that all together, we're down to two variables - we can put everything in terms of train A:

R(A) = 35 = D(A)/T(A)

R(B) = 25 = [500 - D(A)]/T(A)

And since we're solving for T, we can rephrase the first statement as 35T(A) = D(A).

Then plug in 35T(A) for the D(A) in the second equation:

25 = (500 - 35T(A))/T(A) ---> at this point let me drop the (A) since it's all interms of A

25T = 500 - 35T
60T = 500
T = 500/60 = 50/6 = 25/3, or 8 hours and 20 minutes. They started at 1pm, so they'll meet at 9:20.


Now...that's a longer way of doing it, so look at that step where we get to 60T = 500. Since Rate = Distance/Time, if we divide both sides by T we actually replicate just that:

60 (the combined rate) = 500/T

So that's another way of proving that we can add the rates - we get to exactly the same place. And conceptually why we can add the rates is that every mile train A goes, train B doesn't have to. So they're working together, and we can combine their "combined effort" in terms of a combined rate.
Brian Galvin
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Chief Academic Officer
Veritas Prep

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by Sampurn » Thu Oct 06, 2011 12:57 pm
svd.kumar wrote:Hi

Could you please give the answer, for me I got it as 9:20 PM.

Thanks,
SVD
Since relative speed is S1+S2= 35+25 = 60 mph.

=> to cover 60 miles for both the trains, it takes 60 minutes.
=> so, to cover 1 mile, it will take - 60/60 minutes = 1 min.
=> so, to cover 500 miles, it will take 1X500 = 500 minutes = 8 hours, 20 minutes.
=> So, if the train starts at 1 PM, after 8 hours and 20 minutes, they will meet at 9:20 PM. So you are right.