Meeting Times - Trains

[Guest]

The distance from A to B is 31 1/2 miles. The freight train leaves A at 11:30 AM and arrives at B at 12:30 PM. The passenger train leaves B at 11:45 AM and arrives at A at 12:35 PM. When will the trains meet ? The speeds of both trains are uniform.


A ---------------------------------------------------P-----------------------B (Diagram included with problem).


Distance = Rate x Time Rate = Distance / Time Time = Distance / Rate

How do I proceed ?

I don't know what the "P" is.

[Guest]

Either the times you have given are wrong or the speeds aren't the same as one train takes 1 hour to get from A to B, whilst the other takes 50 minutes.

If you only care about when they'll meet (and not where) then the distance between A and B is irrelevant.

If the speeds are the same then it is very easy:

If they started at the same time and it takes them [tex]t[/tex] minutes to make the complete journey, they'll meet exactly in the middle after [tex]t/2[/tex] minutes. Using this fact, we can work out what happens when they don't start at the same time.

Suppose one train starts [tex]s[/tex] minutes after the first. The first train has [tex]t-s[/tex] minutes of journey time left when the second train starts, similarly the second train (because it travels at the same speed) will take [tex]t-s[/tex] minutes to get to where the first train is currently positioned. We are back in the case of two trains taking the same amount of time to make the same journey except the starting and end points aren't A and B, but the positions of the two trains at [tex]s[/tex] minutes.

Using our previous solution we know that they'll meet after [tex](t-s)/2[/tex] minutes. So train 1 will meet train 2, [tex]s+(t-s)/2[/tex] minutes after starting, and train 2 will meet train 1, [tex](t-s)/2[/tex] after starting.

If the speeds aren't the same:

This is a bit more tricky. Suppose train 1 takes [tex]t[/tex] minutes to make the journey, train 2 takes [tex]T[/tex] minutes to make the journey, then if they start at the same time they'll meet after [tex]tT/(t+T)[/tex] minutes (note that if we substitute [tex]T=t[/tex] we get back our previously known result that they meet after [tex]t/2[/tex] minutes).

We can derive [tex]tT/(t+T)[/tex] as follows:
Every minute train 1 covers [tex]1/t[/tex] of the total distance.
Every minute train 2 covers [tex]1/T[/tex] of the total distance.
Every minute the proportion of the total distance covered by both trains combined is [tex]1/t+1/T = (t+T)/(tT)[/tex].
If they meet after [tex]x[/tex] minutes then the proportion of the distance covered is [tex]x(t+T)/(tT)[/tex].
When they meet the proportion covered is [tex]1[/tex], so
[tex]x(t+T)/(tT) = 1[/tex] which rearranges to [tex]x = tT/(t+T)[/tex].

If they start at different times, then things get a bit more complicated yet again. Suppose train 2 starts [tex]s[/tex] minutes after train 1. Train 1 has [tex]t-s[/tex] minutes of its journey left (when train 2 starts), as a proportion this is [tex](t-s)/t[/tex] of its total journey, similarly the distance train 2 has to travel to get to train 1's current position is [tex](t-s)/t[/tex] of its total journey. Let us call [tex]p = (t-s)/t[/tex] to simplify the equations. So after [tex]s[/tex] minutes it will take train 1, [tex]pt[/tex] minutes to get to train 2's current position, and [tex]pT[/tex] minutes for train 2 to get to train 1's current position, so using our previous formula we know they'll meet after:
[tex](pt)(pT)/((pt)+(pT)) = p^2 tT/(p(t+T)) = ptT/(t+T)[/tex] minutes
Substituting [tex]p[/tex], simplifies this to [tex](t-s)T/(t+T)[/tex].

So train 1 will meet train 2, [tex]s+(t-s)T/(t+T)[/tex] minutes after starting. Train 2 will meet train 1, [tex](t-s)T/(t+T)[/tex] minutes after starting.

Hope this helped,

R. Baber.

[Guest]

Ignore my comment in the previous post about the question being wrong, I misread "uniform speed" as "equal speed".

R. Baber.

[Guest]

I am totally confused on both of your solutions.

[Guest]

A simple wee solution........

Question says "when will the trains meet".....I assume that means at what time.......

They will both meet at the same time..... Let the time = t ....the time after 2nd train leaves B....

trainfromA journey time = 1 hours = 31.5 mph (speed)
trainfromB journey time = (1 hours - 10 mins) = 5/6 hours = 0.8333 hour = (31.5 / 0.8333) mph = 37.8 mph (speed)

Speed x time = distance ......

B leaves 15 mins after A, so in 15 mins A travels..... 31.5 x 0.25 = 7.875 miles

dist between trains at 11.45am = 31.5 - 7.875 = 23.625 miles

31.5t + 37.8t = 23.625
69.3t = 23.625
t = 0.341 hours = Trains meet at 20.45 mins after 11:45am = 12:05.45 pm.

[Guest]

Thanks for both solutions.

[Guest]

R. Baber,

Please give more details to your solution. Thanks.

[Guest]

The "simple wee solution" which is fairly well detailed is posted by me.... as Guest.....not R Baber........

[Guest]

My solution is pretty much the same as the other poster's solution, the only real difference is I did it abstractly (without specific times in mind) and I set the distance to 1 unit, instead of 31.5 miles, since it ultimately makes no difference.

[tex]t = 60[/tex] minutes
[tex]T = 50[/tex] minutes
[tex]s = 15[/tex] minutes
So by my solution train 1 meets train 2:
15+(60-15)50/(60+50) = 15+225/11 = 15+20.454545... = 35.45454545... minutes
after 11:30 AM, which is 12:05.4545... PM
Or alternatively
Train 2 meets train 1:
(60-15)50/(60+50) = 225/11 = 20.454545... minutes
after 11:45 AM, which is again 12:05.4545... PM (as expected)

Whereas the other poster first calculated the distance left to travel after 15 minutes, then turned the problem into one where both trains start at the same time. I did it the other way around, and first found the general solution to the problem of both trains leaving at the same time, then applied that solution to the case where the first train travelled for 15 minutes first.

When the other poster found the speed of the trains to be 31.5, and 37.8 miles per hour, I calculate them to be 1/t and 1/T units per minute (I used different units of measurement but it will result in the same ultimate answer).

When the other poster writes
31.5t + 37.8t = 23.625
this is the same as
(1/t+1/T)x = p
in my notation which results in the expression
x = ptT/(t+T)

We get the same answer as each other by pretty much the same method, just the order of certain parts of the calculation are different, and the units of measurement are different.

Hope this helped,

R. Baber.

[Guest]

Thanks again , Guest and R. Baber.

[Guest]

A simple solution....

trainA full journey time = 60 mins......trainB full journey time = 50 mins
TrainB starts 15 mins after trainA, actual time 11:45am
At 11:45am trainA has 15 mins of journey done and 45 mins of journey left to do.
trainA meets trainB at 50/110 x 45 = 20.4545 mins after 11.45am

Isn't that simple.....all I needed was the journey times, the start time of the second train and the actual time.......

How does that work...???

[Guest]

Real Simple.

That works. Thanks again.

[Guest]

Yes, but you didn't answer....How does that work?....Or Why does that work.....

If I mix sand and cement in the ratio of 4:3 there are 4 parts of sand to 3 parts of cement.....that means 7 parts all together.
So sand makes up 4/7 of the mixture and cement makes up 3/7 of the mixture.

It does not seem as straight forward for train times...???

TrainA has journey time of 60 mins and TrainB has journey time of 50 minutes. the ratio of their times is 60:50.
and TrainB is travelling 60/50 times faster than TrainA.

Why do I add their journey times together to get 60 + 50 = 110 ......we cannot just say 110 is the total journey time...???... just same as 4 + 3 = 7 parts sand/cement........ what does the 110 actually mean..??
Also during the time (45 mins) when both trains are moving towards each other I say TrainA meets TrainB at a time 50/110 * 45 = 20.4545
But I cannot seem to say same for TrainB meeting TrainA as 60/110*45 is not equal to 20.4545

? ? ?

[Guest]

I misunderstood your meaning of "How does that work". I thought you meant how does that simple solution work for me. No idea HOW it works.

[Guest]

If TrainA has journey time = t and TrainB has journey time = T

Then following R Baber postings and notation.... an expression for my "simple solution" would be ......

trainA meets trainB at t/(t+T) x (t-s) = 20.4545 mins after 11.45am ......and what is a similar expression viewed from the other train...??

I don't seem to be able to get to this expression using his reasoning.....maybe R Baber can clarify please.

[Guest]

I guess the "symmetric" expression you are after is
[tex](T+s)t/(t+T)[/tex]
from the time train 1 leaves
(instead of [tex](t-s)T/(t+T)[/tex] from the time train 2 leaves).

Instead of thinking in terms of ratios of time (i.e. 60:50) you should be thinking of speed which is effectively the reciprocal of time(e.g. 1/60:1/50).

The expression [tex]tT/(t+T)[/tex] simply comes from the fact that
[tex](1/t)+(1/T)[/tex] = speed of train 1 + speed of train 2 = total speed = 1/"time to meet"
So time to meet = [tex](t^{-1}+T^{-1})^{-1} = tT/(t+T)[/tex]

In order to apply [tex]tT/(t+T)[/tex] you have to have the trains starting at the same time. If they both start at 11.45 you have to multiply by a scale factor of [tex](t-s)/t[/tex] because train 1 has closed the gap during the [tex]s[/tex] minutes. If they both start at 11.30 you have to multiply by a scale factor of [tex](T+s)/T[/tex] because train 2 needed to start from further away so that by the time s minutes have passed it is at B.

Hope this helped,

R. Baber.

[Guest]

Alternatively you could use the expression [tex](t-s)T/(t+T)[/tex] for both trains, if you redefine the variables as follows:
[tex]t =[/tex] time you take
[tex]T =[/tex] time the other guy takes
[tex]s =[/tex] time delay before the other guy starts (can be negative if they start before you)
[tex](t-s)T/(t+T) =[/tex] time you meet after the other guy starts

From train 1's perspective:
[tex]t=60, T=50, s=15,[/tex] they meet [tex](t-s)T/(t+T) = 20.45[/tex] after train 2's start which is 11:45 AM

From train 2's perspective:
[tex]t=50, T=60, s=-15,[/tex] they meet [tex](t-s)T/(t+T) = 35.45[/tex] after train 1's start which is 11:30 AM

Hope this helped,

R. Baber.

[Guest]

Thanks, R Baber for your reply. Yes that explains the "simple solution" better

If trainA journey time = t and trainB journey time = T. trainA leaves 11.30am trainB leaves 11.45am

The "simple solution" trainA meets trainB at t/(t+T) x (t-s) = 20.4545 mins after 11.45am

Now becomes tT/(t + T) x (t - s)/t ...... and the small "t" will cancel leaving my expression

putting in the numbers gives...... (60 x 50)/(60 + 50) x (60 - 15)/60 = 20.4545 mins after 11.45am

simplified gives...... 50/(60 + 50) x (60 - 15) = 50 / 110 x 45 = 20.4545 mins after 11.45am

and viewed from trainB with trainA starting at 11.30am, 15 mins before trainB .....

gives tT/(t + T) x (T + s)/T ... (60 x 50)/(60 + 50) x (50 + 15)/50 = 35.4545 mins after 11.30am

simplifies to T/(t + T) x (T + s) = 60/(60 + 50) x (50 + 15) = 60 / 110 x 65 = 35.4545 mins after 11.30am

While the above solution a ratio and proportion solution..... the ratio of times or the ratio of speeds.....the solution is not the same as would be for the simple example given earlier of a sand/cement mixture.....

sand and cement mixed in ratio of 4:3. means 7 parts all together, so 4/7 of mix is sand and 3/7 of mix is cement.

we have t = 50, T=60, t + T = 110 so one train would be 60/110 x "something" and other train would be 50/110 x "the same something"
defining "the something" in the trains question seems to be the problem....??

PS.
you've beaten me to it with your latest post but I'll post mine anyway...
maybe "the something" is (t-s)/t for both situations with s = +15 if 1st train times are considered and s = -15 if the times of the 2nd train are considered in the expressions.

For the sand/cement situation "the something" is the amount of mixture ..... for the trains, what is it ?.
(t - s)/t is the proportion of the journey still to be completed by the 1st train to move when the 2nd train starts at full speed (uniform).
but what is it when considering it from the 2nd train to move (t- (-s))/t .....??

[Guest]

(t+s)/t is the proportion of the journey left s minutes ago (assuming the train is constantly moving). This can be greater than 1.

You seem to be stuck thinking about ratios, not everything can be resolved with ratios. In this case ratios make no sense. If the trains both leave at the same time one taking t minutes the other T to make the full journey, then even though t is not the same as T when they meet they will both have been travelling for the same amount of time, so how you intend to get this from the equations u(t/(t+T)), and u(T/(t+T)), for some undecided thing u, I don't know. The only thing you can say with ratios, is that the ratios of the speed give you the ratios of the distances travelled (e.g. a train travelling twice as fast will have covered twice the distance compared to its counterpart when they meet). But this is a question about time not distances.

R. Baber.

[Guest]

Thanks for your replies. I only thought this form of solution may be valid when I noticed the "simple" solution 50/110 x 45 gave the answer, similar to the expression for the ratio of components in a mixture. eg. taking the units of time as component parts and since speeds are constant then distances travelled and times would be proportional...........but maybe not in this form.