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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

$\text{Distance}_{\text{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. ??

  • 0
    [Oh, $A$ is $A$ and $B$ is $B$, and never the trains shall meet](http://en.wikipedia.org/wiki/The_Ballad_of_East_and_West)2013-03-19

5 Answers 5

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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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    $T$hank you for such a great explanation and making it so EASY. $T$his definitely makes sense.2012-07-02
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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 $B$.

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simply first train speed=x/4 and second =x/3.5=2x/7 hence first train travels in 2 hours is =x/4*2=x/2 then remaining distance will be x/2

time to meet is = distance/speed distane=x/2 and speed=x/4+2x/7=15x/28 = x/2*28/15x====in hour *60 gives 56 min..............hence 7:56am

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    Please use $\LaTeX$ when typesetting math. Use this guide: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference2013-03-19
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Time taken by the first train 4 hrs. Second train is 3.5hrs. l.c.m. Of 4 & 3.5 = 14 Let 14 be the distance b/w two points(imagine) First train: 4hrs -14km => speed = 3.5km/hr 2nd train : 3.5hrs- 14km => speed = 4km/hr First train has covered 7km in 2hrs At 7a.m the distance b/w two trains is 7km Total speed of two trains-7.5km/hr

60min - 7.5

? -7

=>56min.

Ans : 7.56 am