# When will these two trains meet each other

I cant seem to solve this problem.

A train leaves point A at 5 am and reaches point B at 9 am. Another train leaves point B at 7 am and reaches point A at 10:30 am.When will the two trains meet ? Ans 56 min

Here is where i get stuck. I know that when the two trains meets the sum of their distances travelled will be equal to the total sum , here is what I know so far

Time traveled from A to B by Train 1 = 4 hours

Time traveled from B to A by Train 2 = 7/2 hours

Now if S=Total distance from A To B and t is the time they meet each other then

$$Distance_{Total}= S =\frac{St}{4} + \frac{2St}{7}$$

Now is there any way i could get the value of S so that i could use it here. ??

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The "ans" you wrote makes no sense: 56 minutes...after what?? –  DonAntonio Jul 2 '12 at 0:05
I guess its 56 min after train B leaves the station. So its 7:56 Am –  Rajeshwar Jul 2 '12 at 0:11

We do not need $S$.

The speed of the train starting from $A$ is $S/4$ while the speed of the train starting from $B$ is $S/(7/2) = 2S/7$.

Let the trains meet at time $t$ where $t$ is measured in measured in hours and is the time taken by the train from $B$ when the two trains meet. Note that when train $B$ is about to start train $A$ would have already covered half its distance i.e. a distance of $S/2$.

Hence, the distance traveled by train $A$ when they meet is $\dfrac{S}2 + \dfrac{S \times t}4$.

The distance traveled by train $B$ when they meet is $\dfrac{2 \times S \times t}7$.

Hence, we get that $$S = \dfrac{S}2 + \dfrac{S \times t}{4} + \dfrac{S \times 2 \times t}{7}$$ We can cancel the $S$ since $S$ is non-zero to get $$\dfrac12 = \dfrac{t}4 + \dfrac{2t}7$$ Can you solve for $t$ now? (Note that $t$ is in hours. You need to multiply by $60$ to get the answer in minutes.)

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Thank you for such a great explanation and making it so EASY. This definitely makes sense. –  Rajeshwar Jul 2 '12 at 0:25

Since both trains move toward each other when remaining half of a space, then the general meeting equation in t is: $$v_{1}*t + v_{2}*t = S$$

Then we get that: $$\frac{S t}{4}+\frac{2S t}{7}=\frac{S}{2} \longrightarrow \frac{ t}{4}+\frac{2t}{7}=\frac{1}{2} \longrightarrow t=\frac{14}{15} (\text{56 minutes})$$

Q.E.D.

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Let $d$ be the distance between $A$ and $B$, and assume the trains travel at constant speed. Let $a(t)$ denote the position of the train leaving $A$, and $b(t)$ denote the position of the train leaving $B$. Then we have (assuming $t \in [5,9]$): $$a(t) = \frac{d}{4} (t-5).$$ Similarly for the other train (and remembering that the train is starting at distance $d$, we have (assuming $t \in [7,10.5]$): $$b(t) = d - \frac{d}{3.5} (t-7).$$ To find the time they meet (crash?), we solve for $a(t) = b(t)$, which gives after a minor amount of rearranging (and canceling $d$, which is assumed non-zero), $t = \frac{119}{15}$, which is 7:56 (and lies in $[7,9]$, so the formulae for $a,b$ apply). Thus the trains meet at 7:56, which is 56 minutes after the train departs from $B$.
Please use $\LaTeX$ when typesetting math. Use this guide: meta.math.stackexchange.com/questions/5020/… –  Dennis Gulko Mar 19 '13 at 8:19