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
$$\text{Distance}_{\text{Total}}= S =\frac{St}{4} + \frac{2St}{7} $$ 
Now is there any way i could get the value of S so that i could use it here. ??
 A: 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.)
A: 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.
A: 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 $B$.
A: 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
A: 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
